What is the answer to log_{4}(x-3)+log_{4}(4x+3)=1 ?

The answer to log_{4}(x-3)+log_{4}(4x+3)=1 is x= 13/4

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log_{4}(x-3)+log_{4}(4x+3)=1

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

x = \frac{13}{4}

Solution steps

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

Apply log rules

(x - 3) (4 x + 3) = 4

Solve

(x - 3) (4 x + 3) = 4 : x = \frac{13}{4}, x = - 1

x = \frac{13}{4}, x = - 1

Verify Solutions:

x = \frac{13}{4}

True

, x = - 1

False

The solution is

x = \frac{13}{4}

lo g_{x} (2) = 128

x ln (x) - x = 0

- 1 - ln (2 + x) = 0

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

8 ln (x) = 1

What is the answer to log_{4}(x-3)+log_{4}(4x+3)=1 ?
The answer to log_{4}(x-3)+log_{4}(4x+3)=1 is x= 13/4
Solve for x: log_{4}(x-3)+log_{4}(4x+3)=1
The solution is x= 13/4