What is the answer to log_{8}(x)-log_{x}(64)=1 ?

The answer to log_{8}(x)-log_{x}(64)=1 is x=64,x= 1/8

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log_{8}(x)-log_{x}(64)=1

lo g_{8} (x) - lo g_{x} (64) = 1

x = 64, x = \frac{1}{8}

Decimal Notation

x = 64, x = 0.125

Solution steps

lo g_{8} (x) - lo g_{x} (64) = 1

Apply log rules

lo g_{8} (x) - \frac{2}{lo g _{8} ( x )} = 1

Rewrite the equation with

lo g_{8} (x) = u

u - \frac{2}{u} = 1

Solve

u - \frac{2}{u} = 1 : u = 2, u = - 1

u = 2, u = - 1

Substitute back

u = lo g_{8} (x),

solve for

x

Solve

lo g_{8} (x) = 2 : x = 64

Solve

lo g_{8} (x) = - 1 : x = \frac{1}{8}

x = 64, x = \frac{1}{8}

Verify Solutions:

x = 64

True

, x = \frac{1}{8}

True

The solutions are

x = 64, x = \frac{1}{8}

\frac{ln ( x )}{x} = 3

lo g_{10} (7 - 2 x) = - 1

y = ln (x), x

lo g_{a} (25) + 2 lo g_{a} (625) = 10

lo g_{5} (7 x) - 2 lo g_{5} (3) = 0

What is the answer to log_{8}(x)-log_{x}(64)=1 ?
The answer to log_{8}(x)-log_{x}(64)=1 is x=64,x= 1/8
Solve for x: log_{8}(x)-log_{x}(64)=1
The solution is x=64,x= 1/8