Como resolver este deber de logaritmos

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Hola Elisabeth!

a) log15=log(3·5)=log3+log5=0.47+0.69=1.16

b) log 16= 

$$\begin{align}&log2^4=4log2=4(0.30)=1.20\\&c)\\&log \sqrt{15}= \frac 1 2 log(3·5)= \frac 1 2(log3+log5)=\frac 1 2(0.47+0.69)=1.16\\&\\&d)\\&log12=log(3·2^2)=log3+2log2=0.47+2(0.30)1.07\\&\\&e)\\&log \frac 2 5=log 2-log 5=0.30-0.69=-0.39\\&\\&f)\\&log \frac{15} 2= log 15 - log 2= log(3·5)-log 2=log3+log5-log2=0.86\\&\\&g) log 3^{-3}=-3 log 3=-0.90\\&\\&h) log \sqrt{30}= \frac 1 2 (log10+log3) =\frac 1 2 (log(2·5)+log3)= \frac 1 2(log 2 +log 5 +log 3)=0.73\end{align}$$

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