Question

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

Answer 1

`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!