Resolution of a logarithmic function

[nvez]

Homework Statement

Resolve the following equation:
log₂ x + log₂ [(x-3)/5)] = 3

Homework Equations

Logarithmic laws:
log_c m + log_c n = log_c mn

The Attempt at a Solution

Restriction:

(x-3)/5 > 0
x-3 > 0
x > 3

Resolution:

log₂ x + log₂ [(x-3)/5)] = 3
log₂ x[(x-3)/5)] = 3
log₂ [(x² - 3x)/5x] = 3

2³ = (x² - 3x)/5x
8 = (x² - 3x)/5x
8(5x) = x² - 3x
40x = x² - 3x
40x + 3x = x²
43x = x²
43 = x² / x
43 = x

My answer book says the answer is 8 however I don't seem to get to that after numerous attempts.

Thank you in advanced.

 

[Mark44]

Homework Statement

Resolve the following equation:
log₂ x + log₂ [(x-3)/5)] = 3

Homework Equations

Logarithmic laws:
log_c m + log_c n = log_c mn

The Attempt at a Solution

Restriction:

(x-3)/5 > 0
x-3 > 0
x > 3

Resolution:

log₂ x + log₂ [(x-3)/5)] = 3
log₂ x[(x-3)/5)] = 3

The next line has an error. The denominator should be 5, not 5x.

log₂ [(x² - 3x)/5x] = 3

2³ = (x² - 3x)/5x
8 = (x² - 3x)/5x
8(5x) = x² - 3x
40x = x² - 3x
40x + 3x = x²
43x = x²
43 = x² / x
43 = x

My answer book says the answer is 8 however I don't seem to get to that after numerous attempts.

Thank you in advanced.

 

[nvez]

The next line has an error. The denominator should be 5, not 5x.

Woops! It should have been 5x if it was x/x but it's x/1.

Thank you very much again. :)