Answer
a. $4$
b. $4$
c. $\displaystyle \log_{2}(\frac{32}{2})=\log_{2}32-\log_{2}2$
Work Step by Step
$\log_{b}b^{x}=x$
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a.
Since $16=2^{4},$
$\log_{2}16=\log_{2}2^{4}=4$
b. $ 32=2^{5},\quad 2=2^{1},$ so
$\log_{2}32-\log_{2}2=\log_{2}2^{5}-\log_{2}2^{1}=5-1=4$
c. Conclusion:
$\displaystyle \log_{2}(\frac{32}{2})=\log_{2}32-\log_{2}2$
