What is the answer to solvefor x,3^{2x}-2^{2y}-713=0 ?

The answer to solvefor x,3^{2x}-2^{2y}-713=0 is x=(ln(4^y+713))/(2ln(3))

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solvefor x,3^{2x}-2^{2y}-713=0

solve for

x, 3^{2 x} - 2^{2 y} - 713 = 0

x = \frac{ln ( 4 ^{y} + 713 )}{2 ln ( 3 )}

Solution steps

3^{2 x} - 2^{2 y} - 713 = 0

Move

2^{2 y}

to the right side

3^{2 x} - 713 = 2^{2 y}

Move

713

to the right side

3^{2 x} = 2^{2 y} + 713

Simplify

2^{2 y} + 713 : 4^{y} + 713

3^{2 x} = 4^{y} + 713

Apply exponent rules

2 x ln (3) = ln (4^{y} + 713)

Solve

2 x ln (3) = ln (4^{y} + 713) : x = \frac{ln ( 4 ^{y} + 713 )}{2 ln ( 3 )}

x = \frac{ln ( 4 ^{y} + 713 )}{2 ln ( 3 )}

x, 9^{x} - 2 (3^{x}) - 15 = 0

4^{x - 1} \cdot 5^{2 x + 3} = 6

3^{7 x + 1} = 6561

3^{2 x} - 3^{x} = 6

10 - 2 e^{7 x + 5} = 3

What is the answer to solvefor x,3^{2x}-2^{2y}-713=0 ?
The answer to solvefor x,3^{2x}-2^{2y}-713=0 is x=(ln(4^y+713))/(2ln(3))