Evaluating Logarithms

We will rewrite each expression one at a time using the relationship $${\log }_b(m)=n\iff b^n=m.$$ 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, $$\log (100)=2\iff 10^2=100.$$ The same reasoning applies for the third expression, ${\log }_2(32)=5.$ $${\log }_2(32)=5\iff 2^5=32$$ 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. $$8^2=64\iff {\log }_8(64)=2.$$ For the last expression, this yields $$e^0=1\iff {\log }_e(1)=0\iff \ln (1)=0.$$ To summarize, the following expressions are equivalent. $$\begin{array}{rl}\log (100)=2& \iff 10^2=100\\ 8^2=64& \iff {\log }_8(64)=2\\ {\log }_2(32)=5& \iff 2^5=32\\ e^0=1& \iff \ln (1)=0\end{array}$$

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. $$x^{\text{-}n}=\frac{1}{x^n}.$$ 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, $\log \left(\frac{1}{1000}\right)=\text{-}3.$