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

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

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

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

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

Decimal Notation

x = - 0.41802 \dots

Solution steps

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

Apply log rules

x + 2 = e (x + 1)

Solve

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

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

Verify Solutions:

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

True

The solution is

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

3 lo g_{2} (x + 18) = 12

3 lo g_{10} (x) = lo g_{10} (8)

lo g_{10} (3 x + 1) = 2

lo g_{8} (x) = 2

lo g_{x} (6 x + 16) = 2

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