What is the answer to log_{7}(x-2)+log_{7}(x+4)=1 ?

The answer to log_{7}(x-2)+log_{7}(x+4)=1 is x=3

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log_{7}(x-2)+log_{7}(x+4)=1

lo g_{7} (x - 2) + lo g_{7} (x + 4) = 1

x = 3

Solution steps

lo g_{7} (x - 2) + lo g_{7} (x + 4) = 1

Apply log rules

(x - 2) (x + 4) = 7

Solve

(x - 2) (x + 4) = 7 : x = 3, x = - 5

x = 3, x = - 5

Verify Solutions:

x = 3

True

, x = - 5

False

The solution is

x = 3

3 ln (x - 3) + 4 = 5

2 ln (x + 2) = - 4

lo g_{6} (24 x) - lo g_{6} (2) = 2

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

2.73 = - lo g_{10} (x)

What is the answer to log_{7}(x-2)+log_{7}(x+4)=1 ?
The answer to log_{7}(x-2)+log_{7}(x+4)=1 is x=3
Solve for x: log_{7}(x-2)+log_{7}(x+4)=1
The solution is x=3