Question

How do you solve `log_3 x + log_3 (x - 8) = 2`?

Answer 1

x = 9

Explanation:

Use `log_b m + log_b n = log_b(mn)`

and inverse of `y=log_bx` is `x = b^y`

Here, `log_3 x+log_b(x-8)= log_3(x(x-8))=2`.

Inversely, `x(x-8)=3^2=9`.

So, `x^2-8x-9= (x+1)(x-9)=0`.

`x = -1 and 9`. Negative x is inadmissible for `log_3x`.