What is the answer to log_{2x}(16)=ln(2x) ?

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

Solve for x: log_{2x}(16)=ln(2x)

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

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

x = \frac{e ^{2 l n (2)}}{2}, x = \frac{1}{2 e ^{2 l n (2)}}

Decimal Notation

x = 2.64312 \dots, x = 0.09458 \dots

Solution steps

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

Apply log rules

\frac{4 ln ( 2 )}{ln ( 2 x )} = ln (2 x)

Rewrite the equation with

ln (2 x) = u

\frac{4 ln ( 2 )}{u} = u

Solve

\frac{4 ln ( 2 )}{u} = u : u = 2 ln (2), u = - 2 ln (2)

u = 2 ln (2), u = - 2 ln (2)

Substitute back

u = ln (2 x),

solve for

x

Solve

ln (2 x) = 2 ln (2) : x = \frac{e ^{2 l n (2)}}{2}

Solve

ln (2 x) = - 2 ln (2) : x = \frac{1}{2 e ^{2 l n (2)}}

x = \frac{e ^{2 l n (2)}}{2}, x = \frac{1}{2 e ^{2 l n (2)}}

Verify Solutions:

x = \frac{e ^{2 l n (2)}}{2}

True

, x = \frac{1}{2 e ^{2 l n (2)}}

True

The solutions are

x = \frac{e ^{2 l n (2)}}{2}, x = \frac{1}{2 e ^{2 l n (2)}}

lo g_{3} (x^{2} - 6 x) = 3

2 lo g_{2} (x) + lo g_{2} (a) = 0

6 = - lo g_{10} (x)

lo g_{3} (p) = 6

0.33 = lo g_{10} (x)

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