What is the answer to solvefor x,3^x+2^y=1+2^z ?

The answer to solvefor x,3^x+2^y=1+2^z is x=(ln(1+2^z-2^y))/(ln(3))

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solvefor x,3^x+2^y=1+2^z

solve for

x, 3^{x} + 2^{y} = 1 + 2^{z}

x = \frac{ln ( 1 + 2 ^{z} - 2 ^{y} )}{ln ( 3 )}

Solution steps

3^{x} + 2^{y} = 1 + 2^{z}

Move

2^{y}

to the right side

3^{x} = 1 + 2^{z} - 2^{y}

Apply exponent rules

x ln (3) = ln (1 + 2^{z} - 2^{y})

Solve

x ln (3) = ln (1 + 2^{z} - 2^{y}) : x = \frac{ln ( 1 + 2 ^{z} - 2 ^{y} )}{ln ( 3 )}

x = \frac{ln ( 1 + 2 ^{z} - 2 ^{y} )}{ln ( 3 )}

4 e^{2 x} - 3 = 2

e^{x} \cdot (1 + e^{x}) = \frac{3}{4}

2187 = 2^{x}

1. 5^{x} = 10

1 6^{x} + 2 0^{x} - 2 5^{x} = 0

What is the answer to solvefor x,3^x+2^y=1+2^z ?
The answer to solvefor x,3^x+2^y=1+2^z is x=(ln(1+2^z-2^y))/(ln(3))