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

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

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

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

x = 16

Solution steps

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

Apply log rules

(x - 12) x = 64

Solve

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

x = 16, x = - 4

Verify Solutions:

x = 16

True

, x = - 4

False

The solution is

x = 16

lo g_{2} (2 x + 10) = 5

lo g_{6} (4 x + 5) = 3

lo g_{8} (x + 4) + lo g_{8} (x - 3) = 1

lo g_{x} (x^{2}) = 6

lo g_{10} (x) = 2.1

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