In this section, you will learn to:
- The definition of a logarithmic function as the inverse of an exponential function
- How to write equivalent logarithmic and exponential expressions
- The definitions of the common log and natural log
- Key properties of logarithms
- How to evaluate logarithms using the change of base formula
The Logarithm
Suppose a population of 50 flies is expected to double every week. We can model this with the function \(f(x) = 50(2)^x\), where \(x\) is the number of weeks that have passed. Here's a natural question: when will this population reach 500?
Setting up the equation, we get:
\[500 = 50(2)^x\]
Dividing both sides by 50 to isolate the exponential part gives us:
\[10 = 2^x\]
Now we're stuck. We've built exponential models and used them to make predictions, but we haven't yet talked about how to solve exponential equations — and there's a good reason. None of the algebra tools we've covered so far can handle equations where the variable is in the exponent.
Think about it: we know that \(2^3 = 8\) and \(2^4 = 16\), so \(x\) must be somewhere between 3 and 4 (since \(g(x) = 2^x\) is an increasing function). We could use technology to estimate the answer, but we want an algebraic method to find it exactly.
Context Pause
Why does this matter? In the real world, you'll constantly encounter situations where you need to "undo" exponential growth — figuring out when an investment doubles, how long until a drug leaves your system, or how many years until a population reaches a critical level. Logarithms are the mathematical tool that makes all of this possible.
What we need is an inverse operation for exponentiation — something that "undoes" raising a number to a power. If you've taken algebra before this course, you've seen this: the inverse of an exponential function is a logarithmic function.
We also know that an exponential function has an inverse because it is one-to-one — each output value corresponds to exactly one input value.
Insight Note
Think of logarithms as asking a question. The expression \(\log_2(8)\) is asking: "What power do I raise 2 to in order to get 8?" The answer is 3, because \(2^3 = 8\). A logarithm is just an exponent viewed from the other direction.
If you need to review inverse functions or the one-to-one property, you can refer to your algebra textbook or to College Algebra from OpenStax, available free online at https://openstax.org/details/college-algebra.
Definition 5.3.1: Logarithm
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
The logarithm (base \(b\)) function, written \(\log_b(x)\), is the inverse of the exponential function (base \(b\)), \(b^x\).
\[y = \log_b(x)\]
is equivalent to
\[b^y = x\]
In general, the statement \(b^a = c\) is equivalent to the statement \(\log_b(c) = a\).
Note: The base \(b\) must be positive: \(b > 0\) and \(b \neq 1\).
Definition 5.3.2: Inverse Property of Logarithms
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Since the logarithm and exponential are inverses, it follows that:
\[\log_b(b^x) = x\]
and
\[b^{\log_b(x)} = x\]
These two identities say that logarithms and exponentials "cancel each other out" when composed. They are foundational for solving exponential and logarithmic equations.
Since \(\log\) is a function, it is most correctly written as \(\log_b(c)\), using parentheses to show function evaluation, just as we would with \(f(c)\). However, when the input is a single variable or number, it is common to drop the parentheses and write \(\log_b c\).
Example 5.3.1
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Write these exponential equations as logarithmic equations:
a. \(2^3 = 8\)
b. \(5^2 = 25\)
c. \(10^{-3} = \frac{1}{1000}\)
Example 5.3.1 Solution
To convert from exponential form \(b^a = c\) to logarithmic form \(\log_b(c) = a\), identify the base, the exponent, and the result.
a. \(2^3 = 8\) becomes \(\log_2(8) = 3\)
(Base is 2, exponent is 3, result is 8)
b. \(5^2 = 25\) becomes \(\log_5(25) = 2\)
(Base is 5, exponent is 2, result is 25)
c. \(10^{-3} = \frac{1}{1000}\) becomes \(\log_{10}\left(\frac{1}{1000}\right) = -3\)
(Base is 10, exponent is \(-3\), result is \(\frac{1}{1000}\))
Example 5.3.2
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Write these logarithmic equations as exponential equations:
a. \(\log_6\left(\sqrt{6}\right) = \frac{1}{2}\)
b. \(\log_3(9) = 2\)
Example 5.3.2 Solution
To convert from logarithmic form \(\log_b(c) = a\) to exponential form \(b^a = c\), identify the base, the exponent (the log result), and the argument.
a. \(\log_6\left(\sqrt{6}\right) = \frac{1}{2}\) becomes \(6^{1/2} = \sqrt{6}\)
b. \(\log_3(9) = 2\) becomes \(3^2 = 9\)
By establishing the relationship between exponential and logarithmic functions, we can now solve basic logarithmic and exponential equations by rewriting one form as the other.
Example 5.3.3
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Solve \(\log_4(x) = 2\) for \(x\).
Example 5.3.3 Solution
Rewrite the logarithmic equation in exponential form:
\[4^2 = x\]
So \(x = 16\).
Example 5.3.4
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Solve \(2^x = 10\) for \(x\).
Example 5.3.4 Solution
Rewrite the exponential equation in logarithmic form:
\[x = \log_2(10)\]
While this does define a solution, you might find it a bit unsatisfying — it's hard to compare this expression to the decimal estimate we made earlier. Also, having an exact expression isn't always useful; often we really need a decimal approximation.
The good news: calculators and computers are great at this. The catch: most calculators only evaluate logarithms of two bases — base 10 and base \(e\). But this turns out not to be a problem, because we can use a change of base formula to evaluate logarithms for any base.
Common and Natural Logarithms
Definition 5.3.3: Common Logarithm
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
The common log is the logarithm with base 10, and is typically written \(\log(x)\) (no base shown).
If the base is not indicated in the log function, then the base used is \(b = 10\).
Definition 5.3.4: Natural Logarithm
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
The natural log is the logarithm with base \(e\), and is typically written \(\ln(x)\).
Note: For any other base \(b\) (other than 10), the base must be explicitly written in the notation \(\log_b(x)\).
Context Pause
Why are there two "special" logarithms? Base 10 (\(\log\)) shows up naturally in science and engineering because we use a base-10 number system — the Richter scale for earthquakes and the decibel scale for sound both use common logs. Base \(e\) (\(\ln\)) shows up everywhere in nature — population growth, radioactive decay, compound interest, and statistics all involve the natural logarithm. Your calculator has dedicated buttons for both: LOG and LN.
Example 5.3.5
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Evaluate \(\log(1000)\) using the definition of the common log.
Example 5.3.5 Solution
The following table shows values of the common log for powers of 10:
| Number |
Number as Exponential |
\(\log(\text{number})\) |
| 1000 |
\(10^3\) |
3 |
| 100 |
\(10^2\) |
2 |
| 10 |
\(10^1\) |
1 |
| 1 |
\(10^0\) |
0 |
| 0.1 |
\(10^{-1}\) |
\(-1\) |
| 0.01 |
\(10^{-2}\) |
\(-2\) |
| 0.001 |
\(10^{-3}\) |
\(-3\) |
To evaluate \(\log(1000)\), let:
\[x = \log(1000)\]
Rewrite in exponential form using the common log base of 10:
\[10^x = 1000\]
Since 1000 is the cube of 10, we get:
\[x = 3\]
Alternatively, we can use the inverse property of logs directly:
\[\log_{10}(10^3) = 3\]
Example 5.3.6
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Evaluate \(\log(1/1{,}000{,}000)\).
Example 5.3.6 Solution
Let:
\[x = \log(1/1{,}000{,}000) = \log(1/10^6) = \log(10^{-6})\]
Rewrite in exponential form:
\[10^x = 10^{-6}\]
Therefore \(x = -6\).
Alternatively, using the inverse property of logs:
\[\log_{10}(10^{-6}) = -6\]
Example 5.3.7
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Evaluate:
a. \(\ln e^5\)
b. \(\ln \sqrt{e}\)
Example 5.3.7 Solution
a. To evaluate \(\ln e^5\), let:
\[x = \ln e^5\]
Rewrite in exponential form using the natural log base of \(e\):
\[e^x = e^5\]
Therefore \(x = 5\).
Alternatively, using the inverse property: \(\ln(e^5) = 5\).
b. To evaluate \(\ln \sqrt{e}\), recall that roots can be written as fractional exponents:
\[x = \ln \sqrt{e} = \ln(e^{1/2})\]
Rewrite in exponential form:
\[e^x = e^{1/2}\]
Therefore \(x = \frac{1}{2}\).
Alternatively, using the inverse property: \(\ln(e^{1/2}) = \frac{1}{2}\).
Insight Note
Notice the pattern here: \(\ln(e^{\text{anything}}) = \text{anything}\) and \(\log(10^{\text{anything}}) = \text{anything}\). The logarithm and exponential with the same base always "undo" each other. This is the inverse property in action, and it's one of the most useful tools you'll use when working with logarithms.
Example 5.3.8
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Evaluate the following using your calculator or computer:
a. \(\log 500\)
b. \(\ln 500\)
Example 5.3.8 Solution
a. Using the LOG key on the calculator to evaluate logarithms in base 10:
\[\log(500) \approx 2.69897\]
b. Using the LN key on the calculator to evaluate natural logarithms:
\[\ln(500) \approx 6.21461\]
Some Properties of Logarithms
We often need to evaluate logarithms using a base other than 10 or \(e\). To find a way to use the common or natural logarithm functions to evaluate expressions like \(\log_2(10)\), we need some additional properties.
Definition 5.3.5: Exponent Property of Logarithms
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
\[\log_b(A^q) = q\log_b(A)\]
The exponent property lets you "pull down" an exponent and turn it into a multiplier in front of the logarithm.
Definition 5.3.6: Change of Base Formula
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
\[\log_b(A) = \frac{\log_c(A)}{\log_c(b)} \quad \text{for any bases } b, c > 0\]
The change of base formula lets you convert a logarithm of any base into a ratio of logarithms in a base your calculator can handle (typically base 10 or base \(e\)).
Insight Note
Think of the change of base formula as a "universal translator" for logarithms. Your calculator only speaks two logarithm languages (base 10 and base \(e\)), but with this formula, you can convert any logarithm into one your calculator understands.
Proof of the Exponent Property
We want to show that \(\log_b(A^q) = q\log_b(A)\).
Since the logarithmic and exponential functions are inverses, we can write \(A = b^{\log_b A}\). Therefore:
\[A^q = \left(b^{\log_b A}\right)^q\]
Using the exponential rule \((x^p)^q = x^{pq}\):
\[A^q = b^{q \log_b A}\]
Now take \(\log_b\) of both sides:
\[\log_b(A^q) = \log_b\left(b^{q\log_b A}\right)\]
Using the inverse property on the right side:
\[\log_b(A^q) = q\log_b(A) \quad \checkmark\]
Proof of the Change of Base Property
We want to show that \(\log_b(A) = \frac{\log_c(A)}{\log_c(b)}\) for any bases \(b, c > 0\).
Let \(\log_b(A) = x\). Rewriting in exponential form:
\[b^x = A\]
Take \(\log_c\) of both sides:
\[\log_c(b^x) = \log_c(A)\]
Using the exponent property on the left side:
\[x \cdot \log_c(b) = \log_c(A)\]
Divide both sides by \(\log_c(b)\):
\[x = \frac{\log_c(A)}{\log_c(b)}\]
Since \(x = \log_b(A)\), we have:
\[\log_b(A) = \frac{\log_c(A)}{\log_c(b)} \quad \checkmark\]
Evaluating Logarithms
With the change of base formula, we can finally find a decimal approximation to our question from the beginning of the section.
Example 5.3.9
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Solve \(2^x = 10\) for \(x\).
Example 5.3.9 Solution
First, rewrite the exponential equation as a logarithmic equation:
\[x = \log_2(10)\]
Using the change of base formula, we can convert this to natural logarithms:
\[x = \log_2(10) = \frac{\ln(10)}{\ln(2)}\]
Evaluating with a calculator:
\[x = \frac{\ln(10)}{\ln(2)} \approx 3.3219\]
This finally answers our original question from the beginning of the section: for the population of 50 flies that doubles every week, it will take approximately 3.32 weeks to grow to 500 flies.
Context Pause
This is a powerful result. Before logarithms, there was no algebraic way to solve \(2^x = 10\). You could only guess and check. Logarithms give you an exact answer — and the change of base formula lets you compute it on any calculator. This same technique applies whenever you need to find how long it takes for something that grows (or decays) exponentially to reach a specific level.
Example 5.3.10
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Evaluate \(\log_5(100)\) using the change of base formula.
Example 5.3.10 Solution
We can rewrite this expression using any other base.
Method 1: Using natural logarithm (base \(e\)):
\[\log_5(100) = \frac{\ln(100)}{\ln(5)} \approx 2.861\]
Method 2: Using common logarithm (base 10):
\[\log_5(100) = \frac{\log(100)}{\log(5)} \approx 2.861\]
Both methods give the same answer, confirming that the choice of base \(c\) in the change of base formula doesn't matter — you'll always get the same result.
Summary of Logarithm Properties
Definition 5.3.7: Summary of Logarithm Properties
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Core Definition:
\[y = \log_b(x) \quad \text{is equivalent to} \quad b^y = x\]
In general, \(b^a = c\) is equivalent to \(\log_b(c) = a\).
Note: The base \(b\) must be positive: \(b > 0\) and \(b \neq 1\).
Inverse Properties:
\[\log_b(b^x) = x \quad \text{and} \quad b^{\log_b(x)} = x\]
Exponent Property:
\[\log_b(A^q) = q\log_b(A)\]
Change of Base:
\[\log_b(A) = \frac{\log_c(A)}{\log_c(b)} \quad \text{for any bases } b, c > 0\]
Sum of Logs Property:
\[\log_b(A) + \log_b(C) = \log_b(AC)\]
Difference of Logs Property:
\[\log_b(A) - \log_b(C) = \log_b\left(\frac{A}{C}\right)\]
Log of Reciprocal:
\[\log_b\left(\frac{1}{C}\right) = -\log_b(C)\]
Reciprocal Bases:
\[\log_{1/b}(C) = -\log_b(C)\]
Insight Note
The sum and difference properties are especially useful: they turn multiplication inside a log into addition outside, and division inside into subtraction outside. This is actually the historical reason logarithms were invented — before calculators, scientists used log tables to turn difficult multiplications into simple additions!
Source: The material in this section of the textbook originates from David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, "Chapter 4: Exponential and Logarithmic Functions," licensed under a Creative Commons CC BY-SA 3.0 license. The material here is based on material contained in that textbook but has been modified by Roberta Bloom, as permitted under this license.
Problem Set 5.3
Problem Set 5.3: Logarithms and Logarithmic Functions
Source: Lippman & Rasmussen, Precalculus: An Investigation of Functions (modified by Roberta Bloom)
Rewrite each of these exponential expressions in logarithmic form:
1) \(3^4 = 81\)
Problem 1 Solution
Step 1: Identify the components of the exponential expression \(b^a = c\).
Here \(b = 3\) (base), \(a = 4\) (exponent), and \(c = 81\) (result).
Step 2: Apply the logarithm definition: \(b^a = c\) is equivalent to \(\log_b(c) = a\).
$\(3^4 = 81 \quad \Longleftrightarrow \quad \log_3(81) = 4\)\(
Answer: \)\log_3(81) = 4\(
Verification: Converting back: \)3^4 = 81\( ✓
2) \)10^5 = 100{,}000\(
Problem 2 Solution
Step 1: Identify the components: \)b = 10\(, \)a = 5\(, \)c = 100{,}000\(.
Step 2: Apply the definition \)b^a = c \Longleftrightarrow \log_b(c) = a\(:
\)\(10^5 = 100{,}000 \quad \Longleftrightarrow \quad \log_{10}(100{,}000) = 5\)\(
Answer: \)\log(100{,}000) = 5\( (or equivalently \)\log_{10}(100{,}000) = 5\()
Verification: Converting back: \)10^5 = 100{,}000\( ✓
3) \)5^{-2} = 0.04\(
Problem 3 Solution
Step 1: Identify the components: \)b = 5\(, \)a = -2\(, \)c = 0.04\(.
Step 2: Note that \)0.04 = \frac{1}{25} = \frac{1}{5^2} = 5^{-2}\(, confirming the equation.
Step 3: Apply the definition:
\)\(5^{-2} = 0.04 \quad \Longleftrightarrow \quad \log_5(0.04) = -2\)\(
Answer: \)\log_5(0.04) = -2\(
Verification: Converting back: \)5^{-2} = \frac{1}{25} = 0.04\( ✓
4) \)4^{-1} = 0.25\(
Problem 4 Solution
Step 1: Identify the components: \)b = 4\(, \)a = -1\(, \)c = 0.25\(.
Step 2: Note that \)0.25 = \frac{1}{4} = 4^{-1}\(, confirming the equation.
Step 3: Apply the definition:
\)\(4^{-1} = 0.25 \quad \Longleftrightarrow \quad \log_4(0.25) = -1\)\(
Answer: \)\log_4(0.25) = -1\(
Verification: Converting back: \)4^{-1} = \frac{1}{4} = 0.25\( ✓
5) \)16^{1/4} = 2\(
Problem 5 Solution
Step 1: Identify the components: \)b = 16\(, \)a = \frac{1}{4}\(, \)c = 2\(.
Step 2: Note that \)16^{1/4} = \sqrt[4]{16} = 2\( since \)2^4 = 16\(, confirming the equation.
Step 3: Apply the definition:
\)\(16^{1/4} = 2 \quad \Longleftrightarrow \quad \log_{16}(2) = \frac{1}{4}\)\(
Answer: \)\log_{16}(2) = \frac{1}{4}\(
Verification: Converting back: \)16^{1/4} = \sqrt[4]{16} = 2\( ✓
6) \)9^{1/2} = 3\(
Problem 6 Solution
Step 1: Identify the components: \)b = 9\(, \)a = \frac{1}{2}\(, \)c = 3\(.
Step 2: Note that \)9^{1/2} = \sqrt{9} = 3\(, confirming the equation.
Step 3: Apply the definition:
\)\(9^{1/2} = 3 \quad \Longleftrightarrow \quad \log_9(3) = \frac{1}{2}\)\(
Answer: \)\log_9(3) = \frac{1}{2}\(
Verification: Converting back: \)9^{1/2} = \sqrt{9} = 3\( ✓
---
Rewrite each of these logarithmic expressions in exponential form:
7) \)\log_5(625) = 4\(
Problem 7 Solution
Step 1: Identify the components of the logarithmic expression \)\log_b(c) = a\(.
Here \)b = 5\( (base), \)c = 625\( (argument), \)a = 4\( (result/exponent).
Step 2: Apply the definition: \)\log_b(c) = a\( is equivalent to \)b^a = c\(:
\)\(\log_5(625) = 4 \quad \Longleftrightarrow \quad 5^4 = 625\)\(
Answer: \)5^4 = 625\(
Verification: \)5^4 = 5 \times 5 \times 5 \times 5 = 625\( ✓
8) \)\log_2(1/32) = -5\(
Problem 8 Solution
Step 1: Identify the components: \)b = 2\(, \)c = \frac{1}{32}\(, \)a = -5\(.
Step 2: Apply the definition:
\)\(\log_2\left(\frac{1}{32}\right) = -5 \quad \Longleftrightarrow \quad 2^{-5} = \frac{1}{32}\)\(
Answer: \)2^{-5} = \frac{1}{32}\(
Verification: \)2^{-5} = \frac{1}{2^5} = \frac{1}{32}\( ✓
9) \)\log_{11}(1331) = 3\(
Problem 9 Solution
Step 1: Identify the components: \)b = 11\(, \)c = 1331\(, \)a = 3\(.
Step 2: Apply the definition:
\)\(\log_{11}(1331) = 3 \quad \Longleftrightarrow \quad 11^3 = 1331\)\(
Answer: \)11^3 = 1331\(
Verification: \)11^3 = 11 \times 11 \times 11 = 121 \times 11 = 1331\( ✓
10) \)\log_{10}(0.0001) = -4\(
Problem 10 Solution
Step 1: Identify the components: \)b = 10\(, \)c = 0.0001\(, \)a = -4\(.
Step 2: Apply the definition:
\)\(\log_{10}(0.0001) = -4 \quad \Longleftrightarrow \quad 10^{-4} = 0.0001\)\(
Answer: \)10^{-4} = 0.0001\(
Verification: \)10^{-4} = \frac{1}{10^4} = \frac{1}{10{,}000} = 0.0001\( ✓
11) \)\log_{64}(4) = \frac{1}{3}\(
Problem 11 Solution
Step 1: Identify the components: \)b = 64\(, \)c = 4\(, \)a = \frac{1}{3}\(.
Step 2: Apply the definition:
\)\(\log_{64}(4) = \frac{1}{3} \quad \Longleftrightarrow \quad 64^{1/3} = 4\)\(
Answer: \)64^{1/3} = 4\(
Verification: \)64^{1/3} = \sqrt[3]{64} = 4\( since \)4^3 = 64\( ✓
12) \)\ln(\sqrt{e}) = \frac{1}{2}\(
Problem 12 Solution
Step 1: Identify the components. Recall \)\ln = \log_e\(, so \)b = e\(, \)c = \sqrt{e}\(, \)a = \frac{1}{2}\(.
Step 2: Apply the definition:
\)\(\ln(\sqrt{e}) = \frac{1}{2} \quad \Longleftrightarrow \quad e^{1/2} = \sqrt{e}\)\(
Answer: \)e^{1/2} = \sqrt{e}\(
Verification: By definition, \)e^{1/2} = \sqrt{e}\( ✓
---
If the expression is in exponential form, rewrite it in logarithmic form. If the expression is in logarithmic form, rewrite it in exponential form.
13) \)5^x = 15625\(
Problem 13 Solution
Step 1: This is in exponential form (\)b^a = c\(), so we rewrite in logarithmic form.
Here \)b = 5\(, \)a = x\(, \)c = 15625\(.
Step 2: Apply the definition \)b^a = c \Longleftrightarrow \log_b(c) = a\(:
\)\(5^x = 15625 \quad \Longleftrightarrow \quad \log_5(15625) = x\)\(
Step 3: We can verify the value: \)5^6 = 15625\( (since \)5^2=25\(, \)5^3=125\(, \)5^4=625\(, \)5^5=3125\(, \)5^6=15625\(), so \)x = 6\(.
Answer: \)\log_5(15625) = x\( (and \)x = 6\()
Verification: \)5^6 = 15625\( ✓
14) \)x = 9^3\(
Problem 14 Solution
Step 1: This is in exponential form. Here \)b = 9\(, \)a = 3\(, \)c = x\(.
Step 2: Apply the definition:
\)\(9^3 = x \quad \Longleftrightarrow \quad \log_9(x) = 3\)\(
Step 3: We can compute: \)9^3 = 729\(, so \)x = 729\(.
Answer: \)\log_9(x) = 3\( (equivalently, \)\log_9(729) = 3\()
Verification: \)9^3 = 729\( ✓
15) \)\log_5(125) = x\(
Problem 15 Solution
Step 1: This is in logarithmic form, so we rewrite in exponential form.
Here \)b = 5\(, \)c = 125\(, \)a = x\(.
Step 2: Apply the definition \)\log_b(c) = a \Longleftrightarrow b^a = c\(:
\)\(\log_5(125) = x \quad \Longleftrightarrow \quad 5^x = 125\)\(
Step 3: Since \)5^3 = 125\(, we get \)x = 3\(.
Answer: \)5^x = 125\( (and \)x = 3\()
Verification: \)5^3 = 125\(, so \)\log_5(125) = 3\( ✓
16) \)\log_3(x) = 5\(
Problem 16 Solution
Step 1: This is in logarithmic form. Here \)b = 3\(, \)c = x\(, \)a = 5\(.
Step 2: Rewrite in exponential form:
\)\(\log_3(x) = 5 \quad \Longleftrightarrow \quad 3^5 = x\)\(
Step 3: Compute: \)3^5 = 243\(, so \)x = 243\(.
Answer: \)3^5 = x\( (and \)x = 243\()
Verification: \)3^5 = 243\(, so \)\log_3(243) = 5\( ✓
17) \)\log_{10}(y) = 4\(
Problem 17 Solution
Step 1: This is in logarithmic form. Here \)b = 10\(, \)c = y\(, \)a = 4\(.
Step 2: Rewrite in exponential form:
\)\(\log_{10}(y) = 4 \quad \Longleftrightarrow \quad 10^4 = y\)\(
Step 3: Compute: \)10^4 = 10{,}000\(, so \)y = 10{,}000\(.
Answer: \)10^4 = y\( (and \)y = 10{,}000\()
Verification: \)\log_{10}(10{,}000) = \log_{10}(10^4) = 4\( ✓
18) \)e^x = 10\(
Problem 18 Solution
Step 1: This is in exponential form. Here \)b = e\(, \)a = x\(, \)c = 10\(.
Step 2: Rewrite in logarithmic form using \)\ln = \log_e\(:
\)\(e^x = 10 \quad \Longleftrightarrow \quad \ln(10) = x\)\(
Answer: \)\ln(10) = x\( (or equivalently, \)x = \ln(10) \approx 2.3026\()
Verification: \)e^{\ln(10)} = 10\( by the inverse property ✓
19) \)\ln(x) = -1\(
Problem 19 Solution
Step 1: This is in logarithmic form. Recall \)\ln = \log_e\(, so \)b = e\(, \)c = x\(, \)a = -1\(.
Step 2: Rewrite in exponential form:
\)\(\ln(x) = -1 \quad \Longleftrightarrow \quad e^{-1} = x\)\(
Step 3: Compute: \)e^{-1} = \frac{1}{e} \approx 0.3679\(.
Answer: \)e^{-1} = x\( (and \)x = \frac{1}{e} \approx 0.3679\()
Verification: \)\ln(e^{-1}) = -1\( by the inverse property ✓
20) \)e^5 = y\(
Problem 20 Solution
Step 1: This is in exponential form. Here \)b = e\(, \)a = 5\(, \)c = y\(.
Step 2: Rewrite in logarithmic form:
\)\(e^5 = y \quad \Longleftrightarrow \quad \ln(y) = 5\)\(
Answer: \)\ln(y) = 5\( (and \)y = e^5 \approx 148.413\()
Verification: \)\ln(e^5) = 5\( by the inverse property ✓
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For each equation, rewrite in exponential form and solve for \)x\(.
21) \)\log_5(x) = 3\(
Problem 21 Solution
Step 1: Rewrite the logarithmic equation in exponential form.
Using \)\log_b(c) = a \Longleftrightarrow b^a = c\( with \)b = 5\(, \)a = 3\(, \)c = x\(:
\)\(\log_5(x) = 3 \quad \Longleftrightarrow \quad 5^3 = x\)\(
Step 2: Evaluate \)5^3\(:
\)\(5^3 = 5 \times 5 \times 5 = 125\)\(
Answer: \)x = 125\(
Verification: \)\log_5(125) = \log_5(5^3) = 3\( ✓
22) \)\log_2(x) = -2\(
Problem 22 Solution
Step 1: Rewrite in exponential form with \)b = 2\(, \)a = -2\(, \)c = x\(:
\)\(\log_2(x) = -2 \quad \Longleftrightarrow \quad 2^{-2} = x\)\(
Step 2: Evaluate \)2^{-2}\(:
\)\(2^{-2} = \frac{1}{2^2} = \frac{1}{4} = 0.25\)\(
Answer: \)x = \frac{1}{4}\(
Verification: \)\log_2\left(\frac{1}{4}\right) = \log_2(2^{-2}) = -2\( ✓
23) \)\log_{10}(x) = -3\(
Problem 23 Solution
Step 1: Rewrite in exponential form with \)b = 10\(, \)a = -3\(, \)c = x\(:
\)\(\log_{10}(x) = -3 \quad \Longleftrightarrow \quad 10^{-3} = x\)\(
Step 2: Evaluate \)10^{-3}\(:
\)\(10^{-3} = \frac{1}{10^3} = \frac{1}{1000} = 0.001\)\(
Answer: \)x = \frac{1}{1000} = 0.001\(
Verification: \)\log_{10}(0.001) = \log_{10}(10^{-3}) = -3\( ✓
24) \)\log_3(x) = 6\(
Problem 24 Solution
Step 1: Rewrite in exponential form with \)b = 3\(, \)a = 6\(, \)c = x\(:
\)\(\log_3(x) = 6 \quad \Longleftrightarrow \quad 3^6 = x\)\(
Step 2: Evaluate \)3^6\(:
\)\(3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 9 = 729\)\(
Answer: \)x = 729\(
Verification: \)\log_3(729) = \log_3(3^6) = 6\( ✓
25) \)\log_{25}(x) = \frac{1}{2}\(
Problem 25 Solution
Step 1: Rewrite in exponential form with \)b = 25\(, \)a = \frac{1}{2}\(, \)c = x\(:
\)\(\log_{25}(x) = \frac{1}{2} \quad \Longleftrightarrow \quad 25^{1/2} = x\)\(
Step 2: Evaluate \)25^{1/2}\(. The exponent \)\frac{1}{2}\( means the square root:
\)\(25^{1/2} = \sqrt{25} = 5\)\(
Answer: \)x = 5\(
Verification: \)\log_{25}(5) = \log_{25}(25^{1/2}) = \frac{1}{2}\( ✓
26) \)\log_{64}(x) = \frac{1}{3}\(
Problem 26 Solution
Step 1: Rewrite in exponential form with \)b = 64\(, \)a = \frac{1}{3}\(, \)c = x\(:
\)\(\log_{64}(x) = \frac{1}{3} \quad \Longleftrightarrow \quad 64^{1/3} = x\)\(
Step 2: Evaluate \)64^{1/3}\(. The exponent \)\frac{1}{3}\( means the cube root:
\)\(64^{1/3} = \sqrt[3]{64} = 4 \quad \text{(since } 4^3 = 64\text{)}\)\(
Answer: \)x = 4\(
Verification: \)\log_{64}(4) = \log_{64}(64^{1/3}) = \frac{1}{3}\( ✓
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Evaluate without using your calculator.
27) \)\ln(\sqrt[3]{e})\(
Problem 27 Solution
Step 1: Rewrite the radical as a fractional exponent:
\)\(\ln(\sqrt[3]{e}) = \ln(e^{1/3})\)\(
Step 2: Apply the inverse property \)\ln(e^x) = x\(:
\)\(\ln(e^{1/3}) = \frac{1}{3}\)\(
Answer: \)\frac{1}{3}\(
Verification: \)e^{1/3} = \sqrt[3]{e}\(, and \)\ln(\sqrt[3]{e}) = \frac{1}{3}\( by the inverse property ✓
28) \)\ln\left(\frac{1}{e^2}\right)\(
Problem 28 Solution
Step 1: Rewrite using negative exponents:
\)\(\ln\left(\frac{1}{e^2}\right) = \ln(e^{-2})\)\(
Step 2: Apply the inverse property \)\ln(e^x) = x\(:
\)\(\ln(e^{-2}) = -2\)\(
Answer: \)-2\(
Verification: \)e^{-2} = \frac{1}{e^2}\(, and \)\ln(e^{-2}) = -2\( by the inverse property ✓
29) \)\ln(e^{10})\(
Problem 29 Solution
Step 1: Apply the inverse property \)\ln(e^x) = x\( directly:
\)\(\ln(e^{10}) = 10\)\(
Answer: \)10\(
Verification: The inverse property gives \)\ln(e^{10}) = 10\( directly ✓
30) \)\log_{10}(10^e)\(
Problem 30 Solution
Step 1: Apply the inverse property \)\log_b(b^x) = x\( with \)b = 10\(:
\)\(\log_{10}(10^e) = e\)\(
Note: The exponent \)e \approx 2.71828\( is just a number — the inverse property works for any exponent, not just integers.
Answer: \)e\( (approximately \)2.71828\()
Verification: The inverse property gives \)\log_{10}(10^e) = e\( directly ✓
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For problems 31–38: Evaluate using your calculator. Use the change of base formula if needed.
31) \)\log(20)\(
Problem 31 Solution
Step 1: This is a common logarithm (base 10). Use the LOG key on a calculator:
\)\(\log(20) = \log_{10}(20)\)\(
Step 2: Since \)10^1 = 10\( and \)10^2 = 100\(, we expect the answer between 1 and 2. Evaluating:
\)\(\log(20) \approx 1.3010\)\(
Answer: \)\log(20) \approx 1.3010\(
Verification: \)10^{1.3010} \approx 20\( ✓
32) \)\ln(42)\(
Problem 32 Solution
Step 1: This is a natural logarithm (base \)e\(). Use the LN key on a calculator:
\)\(\ln(42) = \log_e(42)\)\(
Step 2: Since \)e^3 \approx 20.09\( and \)e^4 \approx 54.60\(, we expect the answer between 3 and 4. Evaluating:
\)\(\ln(42) \approx 3.7377\)\(
Answer: \)\ln(42) \approx 3.7377\(
Verification: \)e^{3.7377} \approx 42\( ✓
33) \)\ln(2.9)\(
Problem 33 Solution
Step 1: Use the LN key on a calculator:
\)\(\ln(2.9) = \log_e(2.9)\)\(
Step 2: Since \)e^1 \approx 2.718\( and \)2.9 > e\(, we expect the answer slightly above 1. Evaluating:
\)\(\ln(2.9) \approx 1.0647\)\(
Answer: \)\ln(2.9) \approx 1.0647\(
Verification: \)e^{1.0647} \approx 2.9\( ✓
34) \)\log(0.5)\(
Problem 34 Solution
Step 1: Use the LOG key on a calculator:
\)\(\log(0.5) = \log_{10}(0.5)\)\(
Step 2: Since \)0.5 = \frac{1}{2}\( is between \)10^{-1} = 0.1\( and \)10^0 = 1\(, we expect the answer between \)-1\( and \)0\(. Evaluating:
\)\(\log(0.5) \approx -0.3010\)\(
Answer: \)\log(0.5) \approx -0.3010\(
Verification: \)10^{-0.3010} \approx 0.5\( ✓
35) \)\log_4(36)\(
Problem 35 Solution
Step 1: The base is 4, which is not 10 or \)e\(, so we need the change of base formula:
\)\(\log_b(A) = \frac{\ln(A)}{\ln(b)}\)\(
Step 2: Apply the formula:
\)\(\log_4(36) = \frac{\ln(36)}{\ln(4)}\)\(
Step 3: Evaluate using a calculator:
\)\(\log_4(36) = \frac{\ln(36)}{\ln(4)} = \frac{3.5835}{1.3863} \approx 2.5850\)\(
Answer: \)\log_4(36) \approx 2.5850\(
Verification: \)4^{2.5850} \approx 36\( ✓
36) \)\log_7(100)\(
Problem 36 Solution
Step 1: Use the change of base formula since the base is 7:
\)\(\log_7(100) = \frac{\ln(100)}{\ln(7)}\)\(
Step 2: Evaluate using a calculator:
\)\(\log_7(100) = \frac{\ln(100)}{\ln(7)} = \frac{4.6052}{1.9459} \approx 2.3671\)\(
Answer: \)\log_7(100) \approx 2.3671\(
Verification: \)7^{2.3671} \approx 100\( ✓
37) \)\log_{1.05}(3.5)\(
Problem 37 Solution
Step 1: Use the change of base formula since the base is 1.05:
\)\(\log_{1.05}(3.5) = \frac{\ln(3.5)}{\ln(1.05)}\)\(
Step 2: Evaluate using a calculator:
\)\(\log_{1.05}(3.5) = \frac{\ln(3.5)}{\ln(1.05)} = \frac{1.2528}{0.04879} \approx 25.6773\)\(
This result makes sense in context: if an investment grows at 5% per year (multiplied by 1.05 each year), it takes about 25.68 years for it to grow to 3.5 times its original value.
Answer: \)\log_{1.05}(3.5) \approx 25.6773\(
Verification: \)1.05^{25.6773} \approx 3.5\( ✓
38) \)\log_{1.067}(2)\(
Problem 38 Solution
Step 1: Use the change of base formula since the base is 1.067:
\)\(\log_{1.067}(2) = \frac{\ln(2)}{\ln(1.067)}\)\(
Step 2: Evaluate using a calculator:
\)\(\log_{1.067}(2) = \frac{\ln(2)}{\ln(1.067)} = \frac{0.6931}{0.06488} \approx 10.6833\)\(
This result makes sense in context: if an investment grows at 6.7% per year (multiplied by 1.067 each year), it takes about 10.68 years to double — consistent with the "Rule of 72" estimate of \)72 / 6.7 \approx 10.7\( years.
Answer: \)\log_{1.067}(2) \approx 10.6833\(
Verification: \)1.067^{10.6833} \approx 2$ ✓