# Exponentials and Logarithms I

Alignments to Content Standards: F-BF.B.5

Recall that $\log_b(x)$ is by definition the exponent which $b$ must be raised to in order to yield $x$ ($b > 0$).

Part I

1. Use this definition to compute $\log_{2}(2^5)$.
2. Use this definition to compute $\log_{10}(0.001)$.
3. Use this definition to compute $\ln(e^3)$.
4. Explain why $\log_b(b^y) = y$ where $b>0$.

The above technique can be used to raise numbers to logarithmic powers by first simplifying the exponent.

Part II

1. Evaluate $10^{\log_{10}(100)}.$
2. Evaluate $2^{\log_{2}(\sqrt{2})}.$
3. Evaluate $e^{\ln(89)}.$
4. Explain why $b^{\log_b(x)} = x$ where $b > 0$.

## IM Commentary

The purpose of this task is to help students see the "why" behind properties of logs that are familiar but often just memorized (and quickly forgotten or misremembered). The task focuses on the verbal definition of the log, helping students to concentrate on understanding that a logarithm is an exponent, as opposed to completing a more computational approach.

## Solution

1. Part I
1. $\log_{2}(2^5)=5$ because $5$ is the exponent $2$ must be raised to in order to yield $2^5$.
2. $\log_{10}(0.001) = -3$ because $-3$ is the exponent $10$ is raised to in order to yield $0.001=10^{-3}$.
3. $\ln(e^3)=3$ because $3$ is the exponent $e$ must be raised to in order to yield $e^3$.
4. $\log_b(b^y) = y$ (if $b>0$) because $y$ is the exponent $b$ must be raised to in order to yield $b^y$. In words, when we are finding a logarithm, we are asking "what is the exponent on $b$ in the expression $b^y$?" Under this verbal description of the logarithm, the answer is clearly $y$.

2. Part II

1. From Part I we know that $\log_{10}(100) = \log_{10}(10^2)=2$. So if we raise $10$ to this power, we have $10^{\log_{10}(100)} = 10^2=100$.
2. From Part I we know that $\log_{2}(\sqrt{2})=\log_{2}(2^{\frac12})=\frac12$. So if we raise $2$ to this power, we get $2^{\log_{2}(\sqrt{2})} = 2^{\frac12} =\sqrt{2}$.
3. Thinking through the steps in the last couple of parts, we realize that $e^{\ln(89)} = 89$ because $\ln(89)$ is by definition the exponent $e$ needs to be raised to in order to yield $89$. If we then raise $e$ to exactly this exponent, the result is $89$.
4. By definition, $\log_b(x)$ is the exponent we would raise $b$ to to yield $x$. So if we raise $b$ to that exact exponent, we of course get $x$! This is succinctly stated as the identity$$b^{\log_b(x)} = x \text{ where } b > 0$$