What is the answer to 2+ln(x+1)=0 ?

The answer to 2+ln(x+1)=0 is x= 1/(e^2)-1

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2+ln(x+1)=0

2 + ln (x + 1) = 0

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

Decimal

x = - 0.86466 \dots

Solution steps

2 + ln (x + 1) = 0

Move

2

to the right side

ln (x + 1) = - 2

Apply log rules

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

Solve

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

25 = 10 lo g_{10} (S)

ln (4) + ln (x + 2) = ln (9)

0 = 80 - 9 lo g_{10} (t + 1)

lo g_{27} (x) = \frac{- 1}{3}

lo g_{x} (\frac{1}{2187}) = 7