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

The answer to log_{3}(3x+1)+log_{3}(2)=1 is x= 1/6

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

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

x = \frac{1}{6}

Decimal Notation

x = 0.16666 \dots

Solution steps

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

Move

lo g_{3} (2)

to the right side

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

Apply log rules

3 x + 1 = \frac{3}{2}

Solve

3 x + 1 = \frac{3}{2} : x = \frac{1}{6}

x = \frac{1}{6}

Verify Solutions:

x = \frac{1}{6}

True

The solution is

x = \frac{1}{6}

\frac{1}{2} = lo g_{4} (2 x - 1)

- 1 - ln (x) = 0

x, 0.6 = lo g_{10} (x)

ln (\frac{- 3}{x}) = 0.0633

2.11 = - ln (c)

What is the answer to log_{3}(3x+1)+log_{3}(2)=1 ?
The answer to log_{3}(3x+1)+log_{3}(2)=1 is x= 1/6
Solve for x: log_{3}(3x+1)+log_{3}(2)=1
The solution is x= 1/6