Question

Prescisando dos calculos

Answer 1

a)

$log_{\sqrt8}4=x\\ \\ (\sqrt8)^x=4\\ \\ 2^{\frac{3x}{2}}=2^2\\ \\ \frac{3x}{2}=2\\ \\ 3x=4\\ \\ x=\frac{4}{3}$

b)

$log_{25}0,2=x\\ \\ 25^x=0,2\\ \\ 5^{2x}=\frac{1}{5}\\ \\ 5^{2x}=5^{-1}\\ \\ 2x=-1\\ \\ x=-\frac{1}{2}$