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

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

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

The solution is x=(ln(e^2-1))/(ln(2))

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

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

Decimal Notation

x = 2.67560 \dots

Solution steps

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

Rewrite the equation with

2^{x} = u

ln (1 + u) = 2

Solve

ln (1 + u) = 2 : u = e^{2} - 1

u = e^{2} - 1

Substitute back

u = 2^{x},

solve for

x

Solve

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

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

1 0^{a} = 36.9

3^{x} \cdot 4^{x} = 14 4^{x - 2}

x, e^{x^{2} + x + I n (4)} = 4

150 = 350 - 335 e^{15 x}

55000 = 50000 e^{k \cdot 20}

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