What is the answer to log_{6}(x+4)+log_{6}(x-2)=log_{6}(4x) ?

The answer to log_{6}(x+4)+log_{6}(x-2)=log_{6}(4x) is x=4

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log_{6}(x+4)+log_{6}(x-2)=log_{6}(4x)

lo g_{6} (x + 4) + lo g_{6} (x - 2) = lo g_{6} (4 x)

x = 4

Solution steps

lo g_{6} (x + 4) + lo g_{6} (x - 2) = lo g_{6} (4 x)

Apply log rules

(x + 4) (x - 2) = 4 x

Solve

(x + 4) (x - 2) = 4 x : x = 4, x = - 2

x = 4, x = - 2

Verify Solutions:

x = 4

True

, x = - 2

False

The solution is

x = 4

lo g_{10} (x) = - 2.2

(lo g_{5} (x))^{2} + 5 lo g_{5} (x) + 6 = 0

0.001 = \frac{ln ( x )}{x}

20 x^{3} ln (x) + 9 x^{3} = 0

ln (x) = \frac{1}{3}

What is the answer to log_{6}(x+4)+log_{6}(x-2)=log_{6}(4x) ?
The answer to log_{6}(x+4)+log_{6}(x-2)=log_{6}(4x) is x=4
Solve for x: log_{6}(x+4)+log_{6}(x-2)=log_{6}(4x)
The solution is x=4