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

The answer to ln(x+1)=x-2 is x=-\W_{-1}(-1/(e^3))-1,x=-\W_{0}(-1/(e^3))-1

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

The solution is x=-\W_{-1}(-1/(e^3))-1,x=-\W_{0}(-1/(e^3))-1

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ln (x + 1) = x - 2

x = - W_{- 1} (- \frac{1}{e ^{3}}) - 1, x = - W_{0} (- \frac{1}{e ^{3}}) - 1

Decimal Notation

x = 3.50524 \dots, x = - 0.94753 \dots

Solution steps

ln (x + 1) = x - 2

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

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

Simplify

e^{l n (x + 1)} : x + 1

x + 1 = e^{x - 2}

Solve

x + 1 = e^{x - 2} : x = - W_{- 1} (- \frac{1}{e ^{3}}) - 1, x = - W_{0} (- \frac{1}{e ^{3}}) - 1

x = - W_{- 1} (- \frac{1}{e ^{3}}) - 1, x = - W_{0} (- \frac{1}{e ^{3}}) - 1

Verify Solutions:

x = - W_{- 1} (- \frac{1}{e ^{3}}) - 1

True

, x = - W_{0} (- \frac{1}{e ^{3}}) - 1

True

The solutions are

x = - W_{- 1} (- \frac{1}{e ^{3}}) - 1, x = - W_{0} (- \frac{1}{e ^{3}}) - 1

lo g_{2} (x) - 1 = 3 \cdot lo g_{x} (4)

lo g_{10} (x) = \frac{1}{3}

ln (x) + x = 3

lo g_{2} (a) = lo g_{10} (84)

ln (x - 6) = - 1

What is the answer to ln(x+1)=x-2 ?
The answer to ln(x+1)=x-2 is x=-\W_{-1}(-1/(e^3))-1,x=-\W_{0}(-1/(e^3))-1
Solve for x: ln(x+1)=x-2
The solution is x=-\W_{-1}(-1/(e^3))-1,x=-\W_{0}(-1/(e^3))-1