What is the answer to log_{2}(x)=log_{x}(2)+2 ?

The answer to log_{2}(x)=log_{x}(2)+2 is x=2^{1+sqrt(2)},x=2^{1-sqrt(2)}

Solve for x: log_{2}(x)=log_{x}(2)+2

The solution is x=2^{1+sqrt(2)},x=2^{1-sqrt(2)}

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

x = 2^{1 + 2}, x = 2^{1 - 2}

Solution steps

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

Apply log rules

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

Rewrite the equation with

lo g_{2} (x) = u

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

Solve

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

u = 1 + 2, u = 1 - 2

Substitute back

u = lo g_{2} (x),

solve for

x

Solve

lo g_{2} (x) = 1 + 2 : x = 2^{1 + 2}

Solve

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

x = 2^{1 + 2}, x = 2^{1 - 2}

Verify Solutions:

x = 2^{1 + 2}

True

, x = 2^{1 - 2}

True

The solutions are

x = 2^{1 + 2}, x = 2^{1 - 2}

lo g_{10} (x - 5) = 17

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

lo g_{10} (3 x + 5) = 3

lo g_{10} (x - 5) = - 2

ln (x) = 8.373231552

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