Solve the following equations. (need help)

1) log_x 27 = log_27 x ( read as the logarithm of 27 to base x = logarithm of x to base 27)
2)log_5 x = log_ x 625

Answer : 1) 27 ; 1/27 2) 25 ; 1/25

Can someone please kindly write out the solution of those questions ? i seriously need help. thanks !

1) We need to get both sides to the same log base.

Let y = log_x(27). Then
x^y = x^[log_x(27)] = 27

Take the log base 27 of both sides:

log_27(x^y) = log_27(27)
y log_27(x) = 1
y = 1/log_27(x)

In short, log_x(27) = 1/log_27(x). Put this in the equation:

1/log_27(x) = log_27(x)
1 = [log_27(x)]²
±1 = log_27(x)

log_27(x) = 1 ⇒ x = 27^1 = 27
log_27(x) = -1 ⇒ x = 27^(-1) = 1/27

2) This problem is done in much the same way. The additional complication is that the right side has log_x (625). So we need to manipulate the right side to get something in terms of log_x(5)..

625 = 5*5*5*5 = 5^4. So log_x(625) = log_x(5^4) = 4 log_x(5). Now we can proceed as in the first problem:

log_5(x) = log_x(625)
log_5(x) = 4 log_x(5)
log_5(x) = 4/log_5(x)
log_5(x)]² = 4
log_5(x) = ±2

log_5(x) = 2 ⇒ x = 5^2 = 25
log_5(x) = -2 ⇒ x = 5^(-2) = 1/25