What is the answer to 2log_{6}(x-7)+log_{6}(9)=2 ?

The answer to 2log_{6}(x-7)+log_{6}(9)=2 is x=6^{(2-2log_{6}(3))/2}+7

Solve for x: 2log_{6}(x-7)+log_{6}(9)=2

The solution is x=6^{(2-2log_{6}(3))/2}+7

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2 lo g_{6} (x - 7) + lo g_{6} (9) = 2

x = 6^{\frac{2 - 2 l o g _{6} ( 3 )}{2}} + 7

Solution steps

2 lo g_{6} (x - 7) + lo g_{6} (9) = 2

Move

lo g_{6} (9)

to the right side

2 lo g_{6} (x - 7) = 2 - lo g_{6} (9)

Divide both sides by

2

lo g_{6} (x - 7) = \frac{2 - lo g _{6} ( 9 )}{2}

Apply log rules

x - 7 = 6^{\frac{2 - 2 l o g _{6} ( 3 )}{2}}

Solve

x - 7 = 6^{\frac{2 - 2 l o g _{6} ( 3 )}{2}} : x = 6^{\frac{2 - 2 l o g _{6} ( 3 )}{2}} + 7

x = 6^{\frac{2 - 2 l o g _{6} ( 3 )}{2}} + 7

Verify Solutions:

x = 6^{\frac{2 - 2 l o g _{6} ( 3 )}{2}} + 7

True

The solution is

x = 6^{\frac{2 - 2 l o g _{6} ( 3 )}{2}} + 7

lo g_{8} (x) = 0.333

2 lo g_{3} (x) - lo g_{27} (x) = 10

lo g_{125} (x) = - \frac{1}{3}

y, ln (1 - 2 y) = - 2 t + c

ln (x + 5) - ln (x + 6) = 5

What is the answer to 2log_{6}(x-7)+log_{6}(9)=2 ?
The answer to 2log_{6}(x-7)+log_{6}(9)=2 is x=6^{(2-2log_{6}(3))/2}+7
Solve for x: 2log_{6}(x-7)+log_{6}(9)=2
The solution is x=6^{(2-2log_{6}(3))/2}+7