Algebra 2 Help :)

Question

Log₂x+log₂(x-4)=5. Solve for "x" with examples and show your work

Answer 1

Using the logarithmic product rule (log x*y = log x + log y), we can rewrite

log₂(x) + log₂(x - 4) = log₂(x² - 4x) = 5

Taking the inverse log (log^-1) on both sides (log^-1[log x] = x, log_b^-1 x = b^x)

x² - 4x = 2⁵

x² - 4x - 2⁵ = 0

Solve this for x

x₁ = 8

x₂ = -4 with real numbers, the logarithm is not defined for negative numbers.

Answer 2

log₂x + log₂(x-4) = 5

log₂ [x(x-4)] = 5

x(x-4) = 2⁵

x² - 4x - 32 = 0

(x - 8)(x + 4) = 0

x = 8 or -4

-4 is extraneous (we can't take logs of negative numbers)

So, x = 8