What is the answer to log_{9}(27x)+log_{x}(81)=6 ?

The answer to log_{9}(27x)+log_{x}(81)=6 is x=6561,x=3

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log_{9}(27x)+log_{x}(81)=6

lo g_{9} (27 x) + lo g_{x} (81) = 6

x = 6561, x = 3

Solution steps

lo g_{9} (27 x) + lo g_{x} (81) = 6

Apply log rules

lo g_{9} (27) + lo g_{9} (x) + \frac{2}{lo g _{9} ( x )} = 6

Rewrite the equation with

lo g_{9} (x) = u

lo g_{9} (27) + u + \frac{2}{u} = 6

Solve

lo g_{9} (27) + u + \frac{2}{u} = 6 : u = 4, u = \frac{1}{2}

u = 4, u = \frac{1}{2}

Substitute back

u = lo g_{9} (x),

solve for

x

Solve

lo g_{9} (x) = 4 : x = 6561

Solve

lo g_{9} (x) = \frac{1}{2} : x = 3

x = 6561, x = 3

Verify Solutions:

x = 6561

True

, x = 3

True

The solutions are

x = 6561, x = 3

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

ln (x - 6) + ln (x + 1) = ln (x - 15)

ln (x) + ln (x + 4) = ln (12)

ln (9 x) + 3 = ln (3 x + 2)

lo g_{a} (4) = 1.585

What is the answer to log_{9}(27x)+log_{x}(81)=6 ?
The answer to log_{9}(27x)+log_{x}(81)=6 is x=6561,x=3
Solve for x: log_{9}(27x)+log_{x}(81)=6
The solution is x=6561,x=3