What is the answer to log_{4}(x)+log_{4}(x-12)=3 ?

The answer to log_{4}(x)+log_{4}(x-12)=3 is x=16

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log_{4}(x)+log_{4}(x-12)=3

lo g_{4} (x) + lo g_{4} (x - 12) = 3

x = 16

Solution steps

lo g_{4} (x) + lo g_{4} (x - 12) = 3

Apply log rules

x (x - 12) = 64

Solve

x (x - 12) = 64 : x = 16, x = - 4

x = 16, x = - 4

Verify Solutions:

x = 16

True

, x = - 4

False

The solution is

x = 16

x, ln (y) = 3 ln (x)

ln (x + 3) = 5

2 lo g_{10} (x) = lo g_{10} (36)

lo g_{3} (x + 4) - lo g_{3} (x - 2) = 1

lo g_{3} (- 2 x + 4) = 2

What is the answer to log_{4}(x)+log_{4}(x-12)=3 ?
The answer to log_{4}(x)+log_{4}(x-12)=3 is x=16
Solve for x: log_{4}(x)+log_{4}(x-12)=3
The solution is x=16