What is the answer to ln(y^2-1)-ln(y+1)=3 ?

The answer to ln(y^2-1)-ln(y+1)=3 is y=(e^3+sqrt(e^6-4(-1-e^3)))/2

Solve for y: ln(y^2-1)-ln(y+1)=3

The solution is y=(e^3+sqrt(e^6-4(-1-e^3)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ln (y^{2} - 1) - ln (y + 1) = 3

y = \frac{e ^{3} + e ^{6} - 4 ( - 1 - e ^{3} )}{2}

Decimal Notation

y = 21.08553 \dots

Solution steps

ln (y^{2} - 1) - ln (y + 1) = 3

Add

ln (y + 1)

to both sides

ln (y^{2} - 1) - ln (y + 1) + ln (y + 1) = 3 + ln (y + 1)

Simplify

ln (y^{2} - 1) = 3 + ln (y + 1)

Apply log rules

y^{2} - 1 = e^{3} (y + 1)

Solve

y^{2} - 1 = e^{3} (y + 1) : y = \frac{e ^{3} + e ^{6} - 4 ( - 1 - e ^{3} )}{2}, y = \frac{e ^{3} - e ^{6} - 4 ( - 1 - e ^{3} )}{2}

y = \frac{e ^{3} + e ^{6} - 4 ( - 1 - e ^{3} )}{2}, y = \frac{e ^{3} - e ^{6} - 4 ( - 1 - e ^{3} )}{2}

Verify Solutions:

y = \frac{e ^{3} + e ^{6} - 4 ( - 1 - e ^{3} )}{2}

True

, y = \frac{e ^{3} - e ^{6} - 4 ( - 1 - e ^{3} )}{2}

False

The solution is

y = \frac{e ^{3} + e ^{6} - 4 ( - 1 - e ^{3} )}{2}

x + lo g_{10} (x) = 0

lo g_{3} (4 x) - lo g_{3} (x - 1) = - 3

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

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

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

What is the answer to ln(y^2-1)-ln(y+1)=3 ?
The answer to ln(y^2-1)-ln(y+1)=3 is y=(e^3+sqrt(e^6-4(-1-e^3)))/2
Solve for y: ln(y^2-1)-ln(y+1)=3
The solution is y=(e^3+sqrt(e^6-4(-1-e^3)))/2