What is the answer to log_{6}(x+1)+log_{6}(x+2)=1 ?

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

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log_{6}(x+1)+log_{6}(x+2)=1

lo g_{6} (x + 1) + lo g_{6} (x + 2) = 1

x = 1

Solution steps

lo g_{6} (x + 1) + lo g_{6} (x + 2) = 1

Apply log rules

(x + 1) (x + 2) = 6

Solve

(x + 1) (x + 2) = 6 : x = 1, x = - 4

x = 1, x = - 4

Verify Solutions:

x = 1

True

, x = - 4

False

The solution is

x = 1

lo g_{32} (m) = \frac{1}{5}

lo g_{2} (6 x + 3) = 5

lo g_{5} (x^{4}) + lo g_{5} (x) = 8

lo g_{10} (16) = 2 lo g_{10} (x + 1)

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

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