Evaluating Logarithms

We will rewrite each expression one at a time using the relationship $\log_{b}(m)=n \quad \Leftrightarrow \quad b^n=m.$ Notice that the first and third are written as logarithms and the second and fourth are written as exponents. Since a base is not written on the first logarithm, we know it's base $10.$ Furthermore, $2$ is the exponent. Thus, $\log(100)=2 \quad\Leftrightarrow\quad 10^2=100.$ The same reasoning applies for the third expression, $\log_{2}(32)=5.$ $\log_{2}(32)=5 \quad \Leftrightarrow \quad 2^5=32$ Using the same relationship in the opposite way, we can rewrite the second and fourth expression. In $8^2=64,$ it can be seen that $8$ is the base and $2$ is the exponent. This means that $64$ is the value of which we take the logarithm. $8^2=64 \quad \Leftrightarrow \quad \log_{8}(64)=2.$ For the last expression, this yields $e^0=1 \quad \Leftrightarrow \quad \log_{e}(1)=0 \quad \Leftrightarrow \quad \ln(1)=0.$ To summarize, the following expressions are equivalent. $\begin{aligned} \log(100)=2 &\quad\Leftrightarrow\quad 10^2=100\\ 8^2=64 &\quad\Leftrightarrow\quad \log_8(64)=2\\ \log_2(32)=5 &\quad\Leftrightarrow\quad 2^5=32\\ e^0=1 &\quad\Leftrightarrow\quad \ln(1)=0 \end{aligned}$

When evaluating logarithms, it can be helpful to think about what the expression means. $\log_{3}(81)$ asks the exponent to which $3$ must be raised to equal $81.$ Since, $3 \cdot 3 \cdot 3 \cdot 3 =81,$ $\log_{3}(81)=4.$ The second expression, $\ln(e),$ is a logarithm of base $e.$ It asks the exponent to which $e$ must be raised to equal $e.$ Since $e^1=e,$ $\ln(e)=1.$ Notice that, unlike the other expressions, the last, which is of base $10,$ contains a fraction. Thus, we must consider the exponent to which $10$ is raised to equal $\frac{1}{1000}.$ As it turns out, negative exponents yield fractions. $x^{\text{-} n} = \dfrac{1}{x^n}.$ It can be helpful to rewrite $1000$ as a power of $10.$ Since $1000 = 10^3,$ $\frac{1}{1000} = \frac{1}{10^3} = 10^{\text{-} 3}.$ Thus, $\log \left(\frac{1}{1000} \right)=\text{-} 3.$