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

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

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ln((x+1+e^2)/(x+1))=2

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

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

Decimal

x = 0.15651 \dots

Solution steps

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

Apply log rules

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

Multiply both sides by

x + 1

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

Simplify

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

lo g_{10} (x) = 0.001

x, ln (x) = t

y, x = 20 lo g_{10} (y)

lo g_{2} (7 + x) = 7

x^{l o g_{2} (x)} = 64 x

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