Understand the inverse relationship between exponential and logarithmic functions; convert between exponential and logarithmic form; apply the product, quotient, and power properties of logarithms; and use the natural base e and natural logarithm.

What this topic is asking

The Regents Algebra II exam (the Linear, Quadratic, and Exponential Models, F-LE, and Building Functions, F-BF, clusters) wants you to understand that logarithmic and exponential functions are inverses, to convert between the two forms, to apply the product, quotient, and power properties of logarithms, and to use the natural base $e$ and natural logarithm $\ln$. These functions model growth, decay, and anything that scales multiplicatively.

The inverse relationship

The foundation is that a logarithm is an exponent.

So $\log_2 8 = 3$ because $2^3 = 8$, and $\log_{10} 1000 = 3$ because $10^3 = 1000$. Because the exponential function $b^x$ and the logarithm $\log_b x$ are inverses, applying one undoes the other: $b^{\log_b x} = x$ and $\log_b(b^x) = x$. Converting between exponential and logarithmic form is the move that unlocks most equations.

The properties of logarithms

The log properties come directly from the exponent laws, since a log is an exponent.

To expand, apply them left to right, breaking one log into several. To condense, apply them right to left, combining several into one. A crucial restriction: these act on the argument of the log (the product, quotient, or power inside), never on the outside. So $\log(M + N)$ does not simplify, and $\log M \cdot \log N$ is not $\log(MN)$.

The natural base and natural log

The number $e \approx 2.718$ is the natural base, used for continuous growth and decay. Its inverse is the natural logarithm $\ln x = \log_e x$, so $\ln e = 1$, $\ln 1 = 0$, and $\ln e^x = x$. The natural log obeys the same product, quotient, and power properties as any logarithm. The change-of-base formula, $\log_b M = \frac{\log M}{\log b} = \frac{\ln M}{\ln b}$, lets you evaluate a logarithm of any base on a calculator that only has the common log and the natural log; for instance $\log_2 10 = \frac{\log 10}{\log 2} = \frac{1}{0.3010} \approx 3.32$.

The graph of an exponential function $y = b^x$ (with $b > 1$) rises steeply and has a horizontal asymptote at $y = 0$, passing through $(0, 1)$. Its inverse, the logarithm $y = \log_b x$, is the reflection across the line $y = x$: it rises slowly, has a vertical asymptote at $x = 0$, and passes through $(1, 0)$. Seeing the two graphs as mirror images reinforces why their domains and ranges swap, the exponential's range of positive outputs becomes the logarithm's domain of positive inputs. A clarifying point worth stressing is that the domain of a logarithm is positive numbers only: $\log_b x$ is defined only for $x > 0$, because no power of a positive base gives zero or a negative result. This is why logarithmic equations later require a domain check, and why the graph of a log function has a vertical asymptote at $x = 0$. Keeping the inverse relationship and this domain restriction in mind makes the equation-solving in the next topic feel routine.

Try this

Q1. Rewrite $5^3 = 125$ in logarithmic form. [1 credit]

• Cue. $\log_5 125 = 3$.

Q2. Condense $\log x + \log y$ into one logarithm. [1 credit]

• Cue. Product property: $\log(xy)$.