What is the answer to log_{6}(x+4)+log_{6}(x+3)=1 ?

The answer to log_{6}(x+4)+log_{6}(x+3)=1 is x=-1

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log_{6}(x+4)+log_{6}(x+3)=1

lo g_{6} (x + 4) + lo g_{6} (x + 3) = 1

x = - 1

Solution steps

lo g_{6} (x + 4) + lo g_{6} (x + 3) = 1

Apply log rules

(x + 4) (x + 3) = 6

Solve

(x + 4) (x + 3) = 6 : x = - 1, x = - 6

x = - 1, x = - 6

Verify Solutions:

x = - 1

True

, x = - 6

False

The solution is

x = - 1

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

lo g_{x} (\frac{1}{81}) = - 4

y, x = lo g_{2} (y)

2 x ln (x) - x = 0

8 x^{2} = 64 x \cdot lo g_{2} (x)

What is the answer to log_{6}(x+4)+log_{6}(x+3)=1 ?
The answer to log_{6}(x+4)+log_{6}(x+3)=1 is x=-1
Solve for x: log_{6}(x+4)+log_{6}(x+3)=1
The solution is x=-1