Logarithm help! (one question)?

Question

_5_ indicates 5 as a sub-number. ^2^ indicates 2 as a power-number. In this case, the following equation would read: log base 5 of x-squared = the square of log base 5 of x.


log_5_(x^2^) = [log_5_(x)]^2^

Solve for x. I cannot do it, I admit defeat.

Answer 1

log5(x^2) = log5(x)^2
2 * log5(x) = log5(x) ^ 2
log5(x) ^ 2 - 2 log5(x) = 0

This is a quadratic in log5(x), which we can factor:

log5(x) * ( log5(x) - 2 ) = 0

It has two solutions:

log5(x) = 0, log5(x) = 2

log5(x) = 0 -> x = 5^0 -> x = 1
log5(x) = 2 -> x = 5^2 -> x = 25

Now, let's check the answers:

log5(x^2) = log5(x)^2
log5(1^2) =? log5(1)^2
log5(1) =? 0^2
0 = 0

log5(x^2) = log5(x)^2
log5(25^2) =? log5(25)^2
log5(625) =? 2^2
4 = 4

Answer 2

In the following solution ,log means log to base 5.
log (x²) = (log x)²
2 log x = (log x)²
Let y = log x
2 y = y²
y² - 2y = 0
y (y - 2) = 0
y = 2 is valid solution.
log x = 2
x = 5²
x = 25

Answer 3

log 5 (x^2) - [log 5 (x)]^2 = 0

2 log 5 (x) - [ log 5 (x) ]^2 = 0

log 5 (x) [ 2 - log 5 (x)] = 0

log 5 (x) = 0 or 2 - log 5 (x) = 0

x = 1 or

log 5 (x) = 2 and x = 25