ACT Math: Logarithms on the ACT

Logarithms are a high-value topic on the ACT Math test, appearing in 2-3 questions per exam, typically at medium to high difficulty. Mastering them can be the difference between a good score and a great one, as they test your algebraic manipulation skills and your understanding of inverse relationships.

1. The Foundation: Definition and Conversion

At its core, a logarithm is an exponent. It answers the question: "To what power must I raise the base to get this number?" The formal relationship is expressed in the equation: $${\log }_{b}a=c\text{is equivalent to}{b}^c=a.$$ Here, $b$ is the base, $a$ is the argument, and $c$ is the value of the logarithm.

This equivalence is your most powerful tool. The ACT frequently asks you to convert between logarithmic and exponential forms. For example, if you see ${\log }_{3}81=4$, you can rewrite it as $3^4=81$. Conversely, the equation $10^2=100$ is equivalent to ${\log }_{10}100=2$. A logarithm with base 10, ${\log }_{10}x$, is called the common logarithm and is often written simply as $\log x$ on the ACT. Similarly, ${\log }_{e}x$, where $e\approx 2.718$, is the natural logarithm written as $\ln x$.

Memorizing a few key values accelerates your problem-solving: ${\log }_{b}b=1$ (because $b^1=b$), ${\log }_{b}1=0$ (because $b^0=1$), and ${\log }_b{b}^x=x$.

2. Core Properties for Simplifying Expressions

To simplify logarithmic expressions, you must be fluent with three essential properties. These properties are derived from the laws of exponents and are non-negotiable for the ACT.

1. Product Rule: The log of a product is the sum of the logs.

$${\log }_b(MN)={\log }_{b}M+{\log }_{b}N$$ Example: ${\log }_2(8\cdot 4)={\log }_{2}8+{\log }_{2}4=3+2=5$.

1. Quotient Rule: The log of a quotient is the difference of the logs.

$${\log }_b\left(\frac{M}{N}\right)={\log }_{b}M-{\log }_{b}N$$ Example: ${\log }_5\left(\frac{125}{25}\right)={\log }_{5}125-{\log }_{5}25=3-2=1$.

1. Power Rule: The log of a power allows you to bring the exponent down as a coefficient.

$${\log }_b(M^p)=p\cdot {\log }_{b}M$$ Example: ${\log }_3(81^2)=2\cdot {\log }_{3}81=2\cdot 4=8$.

These properties work in both directions. You can use them to combine several logs into one or to expand a single log into multiple terms. A common ACT task is to express an expression like $2\log x-\frac{1}{2}\log y$ as a single logarithm. Using the power rule first, then the quotient rule, you get: $\log (x^2)-\log (y^{1/2})=\log \left(\frac{x^2}{\sqrt{y}}\right)$.

3. Solving Logarithmic and Exponential Equations

The ACT will present equations where the variable is inside a logarithm or an exponent. Your strategy depends on the setup.

For logarithmic equations like ${\log }_2(x+3)=4$:

1. Convert to exponential form: $2^4=x+3$.
2. Simplify: $16=x+3$.
3. Solve: $x=13$.
4. CRITICAL CHECK: You must ensure the argument $(x+3)$ is positive. Since $13+3>0$, the solution is valid. If your solution made the argument zero or negative, you would discard it.

For exponential equations where the variable is in the exponent, like $5^{x-1}=25$:

1. Rewrite both sides with a common base if possible: $5^{x-1}=5^2$.
2. Since the bases are equal, set the exponents equal: $x-1=2$.
3. Solve: $x=3$.

If you cannot rewrite with a common base, you must use logarithms to "bring down" the exponent. For an equation like $3^x=20$:

1. Take the log (or ln) of both sides: $\log (3^x)=\log (20)$.
2. Apply the Power Rule: $x\cdot \log 3=\log 20$.
3. Solve for $x$: $x=\frac{\log 20}{\log 3}$.

This is a perfectly acceptable final answer on the ACT; you will rarely need to compute a decimal unless asked.

4. Applying the Change of Base Formula

Sometimes you need to evaluate a logarithmic expression with an unfamiliar base, such as ${\log }_{4}7$. Your calculator likely only has $\log$ (base 10) and $\ln$ (base $e$) buttons. The Change of Base Formula is your solution: $${\log }_{b}a=\frac{\log a}{\log b}=\frac{\ln a}{\ln b}.$$ You can use either common log or natural log. To evaluate ${\log }_{4}7$, you would compute $\frac{\log 7}{\log 4}\approx \frac{0.8451}{0.6021}\approx 1.404$.

This formula is also indispensable for solving equations where bases differ, like ${\log }_{3}x={\log }_{9}4$. First, change the base on the right to base 3: ${\log }_{9}4=\frac{{\log }_{3}4}{{\log }_{3}9}=\frac{{\log }_{3}4}{2}$. The equation becomes ${\log }_{3}x=\frac{{\log }_{3}4}{2}$. Use the Power Rule to rewrite the right side as ${\log }_3(4^{1/2})={\log }_{3}2$. Therefore, $x=2$.

Common Pitfalls

1. Misapplying Properties: The rules are specific. ${\log }_b(M+N)$ is NOT ${\log }_{b}M+{\log }_{b}N$. You cannot separate terms inside a log that are being added. Similarly, $\frac{{\log }_{b}M}{{\log }_{b}N}$ is NOT ${\log }_b(M-N)$; it is the Change of Base for ${\log }_{N}M$.

1. Ignoring the Domain: The argument of a logarithm must be strictly positive. When you solve an equation like ${\log }_7(x^2-9)=1$, you get $x^2-9=7^1$, so $x^2=16$, and $x=4$ or $x=-4$. You must check both in the original argument: $4^2-9=7>0$ (good), but $(-4{)}^2-9=7>0$ (also good). Both solutions are valid. If an argument had been zero or negative, that solution would be extraneous.

1. Incorrect Base Handling: When solving $2^{x+1}=8^x$, a student might wrongly set $x+1=8x$. First, you must express both sides with the same base: $2^{x+1}=(2^3{)}^x=2^{3x}$. Now you can set the exponents equal: $x+1=3x$.

1. Overcomplicating Simple Evaluations: Remember your core definitions. A question asking for ${\log }_{8}2$ is not a calculator problem. Ask: "8 to what power is 2?" Since $8^{1/3}=2$, the answer is $\frac{1}{3}$.

Summary

• Logarithms are exponents. The fundamental equivalence is ${\log }_{b}a=c⟺b^c=a$. Mastering this conversion is the first critical step.
• Use the three core properties—Product, Quotient, and Power Rules—to expand, condense, and simplify complex logarithmic expressions efficiently.
• To solve equations, isolate the log or exponential term, then convert between forms. Always check that your solutions yield a positive argument for any remaining logarithms.
• The Change of Base Formula, ${\log }_{b}a=\frac{\log a}{\log b}$, is essential for evaluating logs with any base and for solving equations with different bases.
• On the ACT, these questions test algebraic precision. Work methodically, watch for domain restrictions, and leverage your knowledge of exponents at every step.