What is the answer to x^{1+ln(x)}=e^4*x ?

The answer to x^{1+ln(x)}=e^4*x is x=e^2,x= 1/(e^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{1+ln(x)}=e^4*x

x^{1 + l n (x)} = e^{4} \cdot x

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

Solution steps

x^{1 + l n (x)} = e^{4} x

Apply log rules

(1 + ln (x)) ln (x) = ln (x) + 4

Rewrite the equation with

ln (x) = u

(1 + u) u = u + 4

Solve

(1 + u) u = u + 4 : u = 4, u = - 4

u = 4, u = - 4

Substitute back

u = ln (x),

solve for

x

Solve

ln (x) = 4 : x = e^{2}

Solve

ln (x) = - 4 : x = \frac{1}{e ^{2}}

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

Verify Solutions:

x = e^{2}

True

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

True

The solutions are

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

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

- 3 - ln (- 3 + x) = 0

a^{5} = lo g_{5} (x^{5})

lo g_{X} (16) = - 1

ln (5 x - 4) = ln (8) - ln (x - 2)

What is the answer to x^{1+ln(x)}=e^4*x ?
The answer to x^{1+ln(x)}=e^4*x is x=e^2,x= 1/(e^2)