What is the answer to log_{2}(x-2)+log_{2}(3x+5)=2 ?

The answer to log_{2}(x-2)+log_{2}(3x+5)=2 is x= 7/3

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log_{2}(x-2)+log_{2}(3x+5)=2

lo g_{2} (x - 2) + lo g_{2} (3 x + 5) = 2

x = \frac{7}{3}

Decimal Notation

x = 2.33333 \dots

Solution steps

lo g_{2} (x - 2) + lo g_{2} (3 x + 5) = 2

Apply log rules

(x - 2) (3 x + 5) = 4

Solve

(x - 2) (3 x + 5) = 4 : x = \frac{7}{3}, x = - 2

x = \frac{7}{3}, x = - 2

Verify Solutions:

x = \frac{7}{3}

True

, x = - 2

False

The solution is

x = \frac{7}{3}

lo g_{3} (y) + 3 lo g_{3} (y^{2}) = 2

lo g_{x} (625) = \frac{4}{3}

lo g_{3} (x) = \frac{1}{3}

x, ln (y) = - ln (x)

lo g_{x} (4) = - 2

What is the answer to log_{2}(x-2)+log_{2}(3x+5)=2 ?
The answer to log_{2}(x-2)+log_{2}(3x+5)=2 is x= 7/3
Solve for x: log_{2}(x-2)+log_{2}(3x+5)=2
The solution is x= 7/3