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

The answer to log_{4}(x)=log_{x}(16) is x=4^{sqrt(2)},x= 1/(4^{sqrt(2))}

Solve for x: log_{4}(x)=log_{x}(16)

The solution is x=4^{sqrt(2)},x= 1/(4^{sqrt(2))}

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

x = 4^{2}, x = \frac{1}{4 ^{2}}

Decimal Notation

x = 7.10299 \dots, x = 0.14078 \dots

Solution steps

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

Apply log rules

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

Rewrite the equation with

lo g_{4} (x) = u

u = \frac{2}{u}

Solve

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

u = 2, u = - 2

Substitute back

u = lo g_{4} (x),

solve for

x

Solve

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

Solve

lo g_{4} (x) = - 2 : x = \frac{1}{4 ^{2}}

x = 4^{2}, x = \frac{1}{4 ^{2}}

Verify Solutions:

x = 4^{2}

True

, x = \frac{1}{4 ^{2}}

True

The solutions are

x = 4^{2}, x = \frac{1}{4 ^{2}}

3 ln (5 x + 1) - 8 = 44

ln (x + 1) = ln (x) + ln (1)

lo g_{3} (x - 7) = 4

lo g_{2 x} (16) = - 2

lo g_{2} (x^{2} - 1) - 2 = 0

What is the answer to log_{4}(x)=log_{x}(16) ?
The answer to log_{4}(x)=log_{x}(16) is x=4^{sqrt(2)},x= 1/(4^{sqrt(2))}
Solve for x: log_{4}(x)=log_{x}(16)
The solution is x=4^{sqrt(2)},x= 1/(4^{sqrt(2))}