How do you solve $1/2(log_7x+log_7 8)=log_7 16$ ?

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Answer 1 · 2016-11-26T21:20:13

$log_7 x + log_7 8 = log_7 16/(1/2)$

$log_7 x + log_7 8 = 2log_7 16$

We now apply the rules $log_a n + log_a m = log_a (n xx m)$ and $alogn = logn^a$ .

$log_7(x xx 8) = log_7 16^2$

$log_7 8x = log_7 256$

If $loga = logb$ , then $a = b$ .

$8x = 256$

$x = 32$

Hopefully this helps!

How do you solve 1/2(log_7x+log_7 8)=log_7 161/2(log_7x+log_7 8)=log_7 16?