Topic 2.9 Logarithmic Expressions: define a logarithm as the exponent that produces a given value, and evaluate logarithmic expressions by rewriting them in exponential form.

What this topic is asking

The College Board (Topic 2.9) wants you to understand a logarithm as an exponent: $\log_b y$ is the power to which the base $b$ must be raised to produce $y$. You must convert freely between logarithmic and exponential form, recognize the common log (base $10$) and the natural log (base $e$), and evaluate logarithmic expressions.

A logarithm is an exponent

This equivalence is the entire topic: every logarithmic statement is an exponential statement in disguise, and vice versa.

Converting between forms

The fastest way to evaluate a logarithm is to convert to exponential form. To find $\log_5 125$, ask "$5$ to what power is $125$?" Since $5^3 = 125$, the value is $3$. Going the other way, $2^6 = 64$ becomes $\log_2 64 = 6$. Fluency in both directions is what makes logarithmic evaluation quick on the no-calculator section.

Common and natural logarithms

The natural log is the inverse of $e^x$ and appears throughout continuous-growth modelling; the common log underlies orders of magnitude and the semi-log plots of Topic 2.15.

Why the input must be positive

The domain restriction is not arbitrary. Because a positive base raised to any real exponent is positive, the equation $b^x = y$ has a solution only when $y > 0$. So $\log_b y$ is undefined for $y \leq 0$: there is no power of $2$ that gives $-8$ or $0$. This is the mirror image of the exponential's range being $y > 0$ from Topic 2.3, and it foreshadows the domain $x > 0$ of the logarithmic function in Topic 2.11.

A point worth internalising is that the inverse identities let you cancel a logarithm and an exponential with the same base. Writing $b^{\log_b y} = y$ and $\log_b(b^x) = x$ is the engine behind both evaluating expressions like $\ln e^7$ instantly and solving exponential equations by taking logs. Seeing the logarithm and the exponential as two views of one exponent relationship, rather than as separate machinery, is what makes the rest of the unit feel like one idea instead of many.

Try this

Q1. Evaluate $\log_3 27$. [1 point]

• Cue. $3^x = 27 = 3^3$, so $\log_3 27 = 3$.

Q2. Evaluate $e^{\ln 5}$. [1 point]

• Cue. The inverse identity gives $e^{\ln 5} = 5$.