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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{2x}(16)=log_{e}(2x)

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

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

Solution steps

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

Apply log rules

\frac{4 ln ( 2 )}{ln ( 2 x )} = lo g_{e} (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)}}

2 lo g_{10} (x) + lo g_{10} (4) = 2

- 2 ln (x) - 3 = 0

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

x, ln (x) = 1

ln (y^{2}) = ln (y + 3) + 2

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