How do you solve $2log_b4+log_b5-log_b10=log_bx$ ?

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Answer 1 · 2016-07-17T16:35:34

x=8

Since $nloga=loga^n$ you have:

$log_b 4^2+log_b 5-log_b 10=log_b x$ .

Since $logp+logq=log(pq)$ you have:

$log_b (16*5)-log_b 10=log_b x$ .

Since $logp-logq=log(p/q)$ you have:

$log_b(80/10)=log_b x$ .

Then $log_b8=log_b x$

so x=8

How do you solve 2log_b4+log_b5-log_b10=log_bx2log_b4+log_b5-log_b10=log_bx?