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

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

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

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

b = \frac{1}{2}

Solution steps

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

Move

lo g_{3} (2)

to the right side

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

Apply log rules

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

Solve

b + 1 = \frac{3}{2} : b = \frac{1}{2}

b = \frac{1}{2}

Verify Solutions:

b = \frac{1}{2}

True

The solution is

b = \frac{1}{2}

3^{l o g_{x} (81)} = x

lo g_{x} (625) = - 4

lo g_{9} (2 x) - lo g_{3} (2) = 1

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

lo g_{2} (x) = 10000000

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