Đáp án:
$log_{49}32=\dfrac{5}{2(a-1)}$
Giải thích các bước giải:
$log_{49}32=log_{7^2}2^5$
$=\dfrac{5}{2}log_72$
$=\dfrac{5}{2}.\dfrac{1}{log_27}$
$=\dfrac{5}{2}.\dfrac{1}{log_27.2-1}$
$=\dfrac{5}{2}.\dfrac{1}{log_214-1}$
$=\dfrac{5}{2}.\dfrac{1}{a-1}$
$=\dfrac{5}{2(a-1)}$
