Question

log base 4/5 ×64/125 +logbase4/5 ×1​

Answer 1

Answer:

Step-by-step explanation:

I think this is correct and helpful

Answer 2

Answer:

3

Explanation:

$log_{\frac{4}{5}}\frac{64}{125}+log_{\frac{4}{5}}1$

= $log_{\frac{4}{5}}\frac{4^{3}}{5^{3}}+0$

=$log_{\frac{4}{5}}(\frac{4}{5})^{3}$

= $3$
$\boxed {log_{a}a=1}$

$\boxed{loga^{m} = mloga}$

••••