How do you solve this equation for x

(1/4)^(2x-3) = 8^(x+5)

2x-3 is the exponent of 1/4 and x+5 is the exponent of 8.

2^(-4x+6) = 2^(3x+15) , now solve for x

Apply ln!

(2*x - 3)*ln(1/4) = (x + 5)*ln(8)

2*x*ln(1/4) - 3*ln(1/4) = x*ln(8) + 5*ln(8)

2*x*ln(1/4) - x*ln(8) = 5*ln(8) + 3*ln(1/4)

2*x = (5*ln(8) + 3*ln(1/4))/(2*ln(1/4) - ln(8))

x = (5*ln(8) + 3*ln(1/4))/(2*ln(1/4) - ln(8))

x = - 1.285714286

...both (1/4) and 8 are base-2 numbers

(1/4) = 2^(- 2)

8 = 2^3

so,

[2^(- 2)]^(2x - 3) = 2^(- 4x + 6) = 2^(3x + 15)

so...

- 4x + 6 = 3x + 15

7x = - 9
x = - 9/7

check

qed

¼^(2x - 3) = 8^(x + 5)
(2x - 3) Log ¼ = (x + 5) Log 8
(2x - 3)(- 0.6021) = (x + 5)(0.9031)
- 1.21x + 1.81 = 0.9031x + 4.52
- 1.21x - 0.9031x = 4.52 - 1.81
- 2.1x = 2.71
x = 2.71 / - 2.1
x = - 1.29
¯¯¯¯¯¯¯¯¯

1/4 = 8^(-2/3), so

(8^(-2/3))^(2x-3) = 8^(x+5), or
8^(-2/3*(2x-3)) = 8^(x+5)

I think you can figure out the rest.