Compute all values of xx that satisfy (log4x2−4)3+(log8x3−8)3=(log2x2−12)3.(log4⁡x2−4)3+(log8⁡x3−8)3=(log2⁡x2−12)3.?

Answer 1

The equation isn't clear as to exponents and there are 2 equals signs, but this is what I assume it to be:

(log₄x² - 4)³ + (log₈x³ - 8)³ = (log₂x² - 12)³

By the laws of logs: (2 log₄x - 4)³ + (3 log₈x - 8)³ = (2 log₂x -(2y - 12 ) 12)³ . . . . . . . . . eqn (1)

Now, convert all to log₂: log₄x = log₂x / log₂4 = log₂x / 2

and log₈x = log₂x / log₂8 = log₂x / 3

Eqn (1) becomes:

(2 log₂x / 2 - 4)³ + (3 log₂x / 3 - 8)³ = (2 log₂x - 12)³

(log₂x - 4)³ + (log₂x - 8)³ = (2 log₂x - 12)³

Let y = log₂x: (y - 4)³ + (y - 8)³ = (2y - 12)³

I leave it to you to multiply out the 3 cubed terms in parentheses, group powers of y, solve for y, and substitute back for x.

Answer 2

Assuming you mean:

log(4x² - 4)³ - log(8x² - 8)³ = log(2x² - 13)³

3log(4x² - 4) - 3log(8x² - 8) = 3log(2x² - 13)

log(4x² - 4) - log(8x² - 8) = log(2x² - 13)

log[(4x² - 4) / (8x² - 8)] = log(2x² - 13)

(4x² - 4) / (8x² - 8) = 2x² - 13

Can you solve for x and answer?