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

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

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

solve for

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

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

Solution steps

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

Move

3^{y}

to the right side

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

Apply exponent rules

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

Solve

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

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

2^{2 x - 1} = 5^{3 - x}

4^{x + 3} = 6^{x}

16 - 5 \cdot 2^{x} + 4^{x - 1} = 0

9^{x} = 3^{x - 1}

x, 4^{x} = 10 - 4^{x + 1}

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