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

The answer to ln(x+1)=1+ln(x-1) is x=(e+1)/(e-1)

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ln(x+1)=1+ln(x-1)

ln (x + 1) = 1 + ln (x - 1)

x = \frac{e + 1}{e - 1}

Decimal Notation

x = 2.16395 \dots

Solution steps

ln (x + 1) = 1 + ln (x - 1)

Apply log rules

x + 1 = e (x - 1)

Solve

x + 1 = e (x - 1) : x = \frac{e + 1}{e - 1}

x = \frac{e + 1}{e - 1}

Verify Solutions:

x = \frac{e + 1}{e - 1}

True

The solution is

x = \frac{e + 1}{e - 1}

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

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

lo g_{2} (8 x - 3) = lo g_{10} (200)

(ln (x))^{2} - 2 ln (x^{3}) = - 5

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

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