What is the answer to log_{4}(3x-4)+log_{4}(x+3)=2 ?

The answer to log_{4}(3x-4)+log_{4}(x+3)=2 is x= 7/3

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log_{4}(3x-4)+log_{4}(x+3)=2

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

x = \frac{7}{3}

Decimal Notation

x = 2.33333 \dots

Solution steps

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

Apply log rules

(3 x - 4) (x + 3) = 16

Solve

(3 x - 4) (x + 3) = 16 : x = \frac{7}{3}, x = - 4

x = \frac{7}{3}, x = - 4

Verify Solutions:

x = \frac{7}{3}

True

, x = - 4

False

The solution is

x = \frac{7}{3}

4.4 = - lo g_{10} (x)

lo g_{10} (x (20 - x)) = 2

\frac{ln ( p )}{ln ( 5 )} - 4 = 0

ln (\frac{x - 5}{x - 1}) = ln (6)

0.27 = ln (x)

What is the answer to log_{4}(3x-4)+log_{4}(x+3)=2 ?
The answer to log_{4}(3x-4)+log_{4}(x+3)=2 is x= 7/3
Solve for x: log_{4}(3x-4)+log_{4}(x+3)=2
The solution is x= 7/3