How do you solve $log_b30-log_b5^2=log_bx$ ?

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Answer 1 · 2016-07-25T21:22:57

$x=6/5$

$log_b 30-log_b 5^2=log_b x ?$

$"so ;" log _a( b/c)=log_a b-log_a c$

$log _b (30/25)=log _b x$

$30/25=x$

$x=6/5$

How do you solve log_b30-log_b5^2=log_bxlog_b30-log_b5^2=log_bx?