What is the answer to log_{4}(x)-log_{x}(16)=1 ?

The answer to log_{4}(x)-log_{x}(16)=1 is x=16,x= 1/4

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log_{4}(x)-log_{x}(16)=1

lo g_{4} (x) - lo g_{x} (16) = 1

x = 16, x = \frac{1}{4}

Decimal Notation

x = 16, x = 0.25

Solution steps

lo g_{4} (x) - lo g_{x} (16) = 1

Apply log rules

lo g_{4} (x) - \frac{2}{lo g _{4} ( x )} = 1

Rewrite the equation with

lo g_{4} (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_{4} (x),

solve for

x

Solve

lo g_{4} (x) = 2 : x = 16

Solve

lo g_{4} (x) = - 1 : x = \frac{1}{4}

x = 16, x = \frac{1}{4}

Verify Solutions:

x = 16

True

, x = \frac{1}{4}

True

The solutions are

x = 16, x = \frac{1}{4}

x, 7.09 = - lo g_{10} (x)

ln (x^{4}) = 4 ln (x)

ln (x + 1) - 3 = 0

6.3 = lo g_{10} (x)

lo g_{10} (lo g_{10} (a)) = 0.7

What is the answer to log_{4}(x)-log_{x}(16)=1 ?
The answer to log_{4}(x)-log_{x}(16)=1 is x=16,x= 1/4
Solve for x: log_{4}(x)-log_{x}(16)=1
The solution is x=16,x= 1/4