Solving logarithmic equation log2 (9∧x-4)=log2 (3∧x-2)+3

Solving logarithmic equation log2 (9∧x-4)=log2 (3∧x-2)+3

log2(9^x-4)=log2(3^x-2)+log2(8)
log2(3^2x-4)=log2(8×3^x-8×2)
3^2x-4=8×3^x-16
Let a = 3 ^ X
a^2-4=8a-16
a^2-8a+12=0
a1=2,a2=6
∴3^x1=2,3^x2=6
x1=log(3)2
x2=log(3)6

Logarithmic equation
Log2x + 2log4x + log8x = 7 find X.. (where the number after log is the base...)

7 = log(2)x + 2log(4)x + log(8)x
= log(2)x + 2[log(2)x]/2 + log(2)x/3
= (7/3)log(2)x,
log(2)x = 3
x = 2^3 = 8

The logarithmic equation is solved
log5x(5/x)+(log5(x))^2=1

It's not made up, is it
It seems that it's not made up
Answer: x = 1,5,1 / 25
log5x(5/x)=1-(log5(x))^2=(1-log5(x))*(1+log5(x))=(log5(5/x))*(log5(5x)
Bottom changing formula
lg(5/x)/lg(5x)=lg(5/x)*lg(5x)/(lg5)^2
If LG (5 / x) = 0, then x = 5
Otherwise, (LG (5)) ^ 2 = (LG (5x)) ^ 2
LG (5x) = LG (5) or LG (5x) = - LG (5)
X = 1 or x = 1 / 25