Question

What is `x` in the equation `log_3x^2 - log_"5"25 = log_"2"16`?

Answer 1

`x = +-27`

Explanation:

`log_5 25 = log_5 5^2 = 2`

`log_2 16 = log_2 2^4 = 4`

So:

`log_3 x^2 = log_2 16 + log_5 25 = 4+2=6`

Hence:

`x^2 = 3^(log_3 x^2) = 3^6`

So

`x = +-sqrt(3^6) = +-3^3 = +-27`

Footnote

Notice that: `log_3 x^2 = 2 log_3 abs(x)` and not simply `2 log_3 x`.

If I had thought that `log_3 x^2 = 2 log_3 x`, then I would only have found the solution `x=27`.