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

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

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

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

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

Decimal

x = 0.15651 \dots

Solution steps

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

Apply log rules

x + 1 = e^{2} x

Solve

x + 1 = e^{2} x : 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}

x, 5 x + x ln (x) = 0

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

ln (3 x) = 2 x - 5

- 1 = 20 lo g_{10} (x)

lo g_{10} (x) - lo g_{10} (x + 3) = - 1

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