What is the answer to ln(x^2+2)-ln(x-1)=2 ?

The answer to ln(x^2+2)-ln(x-1)=2 is x=(e^2+sqrt(e^4-4(2+e^2)))/2 ,x=(e^2-sqrt(e^4-4(2+e^2)))/2

Solve for x: ln(x^2+2)-ln(x-1)=2

The solution is x=(e^2+sqrt(e^4-4(2+e^2)))/2 ,x=(e^2-sqrt(e^4-4(2+e^2)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ln (x^{2} + 2) - ln (x - 1) = 2

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

Decimal Notation

x = 5.75862 \dots, x = 1.63043 \dots

Solution steps

ln (x^{2} + 2) - ln (x - 1) = 2

Add

ln (x - 1)

to both sides

ln (x^{2} + 2) - ln (x - 1) + ln (x - 1) = 2 + ln (x - 1)

Simplify

ln (x^{2} + 2) = 2 + ln (x - 1)

Apply log rules

x^{2} + 2 = e^{2} (x - 1)

Solve

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

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

Verify Solutions:

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}

True

, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

True

The solutions are

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

15 = 6 lo g_{4} (8 x)

ln (x) - \frac{3}{ln ( x )} = 2

lo g_{3} (x + 17) - 2 = lo g_{3} (2 x)

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

2 lo g_{2} (x) = 4

What is the answer to ln(x^2+2)-ln(x-1)=2 ?
The answer to ln(x^2+2)-ln(x-1)=2 is x=(e^2+sqrt(e^4-4(2+e^2)))/2 ,x=(e^2-sqrt(e^4-4(2+e^2)))/2
Solve for x: ln(x^2+2)-ln(x-1)=2
The solution is x=(e^2+sqrt(e^4-4(2+e^2)))/2 ,x=(e^2-sqrt(e^4-4(2+e^2)))/2