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

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

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

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

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ln (x + 1) = 2 + ln (x - 1)

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

Decimal Notation

x = 1.31303 \dots

Solution steps

ln (x + 1) = 2 + ln (x - 1)

Apply log rules

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

lo g_{2} (y) - 8 \cdot lo g_{y} (2) = 2

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

lo g_{2} (x + 5) - lo g_{2} (x - 1) = 2

b, lo g_{3} (\frac{a}{b}) = lo g_{27} (a) + 1

lo g_{9} (x - 5) + lo g_{9} (x + 3) = 1

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