What is the answer to log_{6}(8)+log_{6}(x-7)=1 ?

The answer to log_{6}(8)+log_{6}(x-7)=1 is x= 31/4

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log_{6}(8)+log_{6}(x-7)=1

lo g_{6} (8) + lo g_{6} (x - 7) = 1

x = \frac{31}{4}

Solution steps

lo g_{6} (8) + lo g_{6} (x - 7) = 1

Move

lo g_{6} (8)

to the right side

lo g_{6} (x - 7) = 1 - lo g_{6} (8)

Apply log rules

x - 7 = \frac{3}{4}

Solve

x - 7 = \frac{3}{4} : x = \frac{31}{4}

x = \frac{31}{4}

Verify Solutions:

x = \frac{31}{4}

True

The solution is

x = \frac{31}{4}

lo g_{10} (x^{2}) + 3 x = 1

u, lo g_{3} (3 u) = t

lo g_{5} (2 x - 3) = 0

2 lo g_{10} (x) = lo g_{10} (4 x + 5)

lo g_{10} (x^{2} - 1) = 2

What is the answer to log_{6}(8)+log_{6}(x-7)=1 ?
The answer to log_{6}(8)+log_{6}(x-7)=1 is x= 31/4
Solve for x: log_{6}(8)+log_{6}(x-7)=1
The solution is x= 31/4