## Intermediate Algebra for College Students (7th Edition)

$\dfrac{1}{2}$
Note that $10=\sqrt{100}$. Replace $10$ with its equivalent of $\sqrt{100}$ to obtain: $=\log_{100}{(\sqrt{100})}$ Use the rule $\sqrt{a}= a^{\frac{1}{2}}$ to obtain: $=\log_{100}{(100^{\frac{1}{2}})}$ RECALL: $\log_b{b^x} = x$ Use the rule above to obtain: $=\dfrac{1}{2}$