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

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

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

The solution is x=(e^2+sqrt(e^4-4(-1+e^2)))/2

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

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

Decimal Notation

x = 6.38905 \dots

Solution steps

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

Add

ln (x - 1)

to both sides

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

Simplify

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

Apply log rules

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

Solve

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

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

Verify Solutions:

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

True

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

False

The solution is

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

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

lo g_{8} (16) + lo g_{8} (x - 2) = 2

0.7 = - lo g_{10} (x)

x^{l o g_{10} (x)} = 1 0^{6} x

y, ln (y) = 3 + ln (x)

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