What is the answer to log_{3}(3x)=log_{x}(81)+1 ?

The answer to log_{3}(3x)=log_{x}(81)+1 is x=9,x= 1/9

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log_{3}(3x)=log_{x}(81)+1

lo g_{3} (3 x) = lo g_{x} (81) + 1

x = 9, x = \frac{1}{9}

Decimal Notation

x = 9, x = 0.11111 \dots

Solution steps

lo g_{3} (3 x) = lo g_{x} (81) + 1

Apply log rules

lo g_{3} (3) + lo g_{3} (x) = \frac{4}{lo g _{3} ( x )} + 1

Rewrite the equation with

lo g_{3} (x) = u

lo g_{3} (3) + u = \frac{4}{u} + 1

Solve

lo g_{3} (3) + u = \frac{4}{u} + 1 : u = 2, u = - 2

u = 2, u = - 2

Substitute back

u = lo g_{3} (x),

solve for

x

Solve

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

Solve

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

x = 9, x = \frac{1}{9}

Verify Solutions:

x = 9

True

, x = \frac{1}{9}

True

The solutions are

x = 9, x = \frac{1}{9}

ln (2 x - 1) = 8

lo g_{x} (\frac{1}{100}) = - 2

x e^{l n (x)} - 6 = 6 - x

lo g_{x} (2) + lo g_{2} (x) = 2.5

lo g_{x} (8) = 0.5

What is the answer to log_{3}(3x)=log_{x}(81)+1 ?
The answer to log_{3}(3x)=log_{x}(81)+1 is x=9,x= 1/9
Solve for x: log_{3}(3x)=log_{x}(81)+1
The solution is x=9,x= 1/9