How do you find the exact value of $log_8 32+log_8 2$ ?

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Answer 1 · 2017-12-11T16:24:13

$2$ .

Recall that, $log_bm+log_bn=log_b(mn).$

$:. log_8 32+log_8 2,$

$=log_8 (32xx2),$

$=log_8 64,$

$=log_8 8^2,$

$=2log_8 8,........[because, log_b(m^n)=nlog_bm],$

$=2xx1.................[because, log_b b=1].$

$rArr log_8 32+log_8 2=2$ .

How do you find the exact value of log_8 32+log_8 2log_8 32+log_8 2?