What is the answer to log_{3}(x-1)+log_{3}(2x+1)=2 ?

The answer to log_{3}(x-1)+log_{3}(2x+1)=2 is x= 5/2

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log_{3}(x-1)+log_{3}(2x+1)=2

lo g_{3} (x - 1) + lo g_{3} (2 x + 1) = 2

x = \frac{5}{2}

Decimal Notation

x = 2.5

Solution steps

lo g_{3} (x - 1) + lo g_{3} (2 x + 1) = 2

Apply log rules

(x - 1) (2 x + 1) = 9

Solve

(x - 1) (2 x + 1) = 9 : x = \frac{5}{2}, x = - 2

x = \frac{5}{2}, x = - 2

Verify Solutions:

x = \frac{5}{2}

True

, x = - 2

False

The solution is

x = \frac{5}{2}

lo g_{7} (5 x - 6) = 3

lo g_{5} (2 x - 23) = lo g_{5} (x + 51)

ln (1 - x) = 0

ln (x^{2}) = - 2

- 1 - 5 \cdot ln (x) = 0.5

What is the answer to log_{3}(x-1)+log_{3}(2x+1)=2 ?
The answer to log_{3}(x-1)+log_{3}(2x+1)=2 is x= 5/2
Solve for x: log_{3}(x-1)+log_{3}(2x+1)=2
The solution is x= 5/2