What is the answer to solvefor x,z^2=3^{10}+3^x ?

The answer to solvefor x,z^2=3^{10}+3^x is x=(ln(z^2-59049))/(ln(3))

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solvefor x,z^2=3^{10}+3^x

solve for

x, z^{2} = 3^{10} + 3^{x}

x = \frac{ln ( z ^{2} - 59049 )}{ln ( 3 )}

Solution steps

z^{2} = 3^{10} + 3^{x}

Switch sides

3^{10} + 3^{x} = z^{2}

Move

3^{10}

to the right side

3^{x} = z^{2} - 3^{10}

Simplify

3^{x} = z^{2} - 59049

Apply exponent rules

x ln (3) = ln (z^{2} - 59049)

Solve

x ln (3) = ln (z^{2} - 59049) : x = \frac{ln ( z ^{2} - 59049 )}{ln ( 3 )}

x = \frac{ln ( z ^{2} - 59049 )}{ln ( 3 )}

3 (1 2^{- 11 - x}) + 8 = 2501

e^{x} - e^{- x} - 6 = - 10

3^{x} + 20 = 28

y, e^{x + 2 y} = 0

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

What is the answer to solvefor x,z^2=3^{10}+3^x ?
The answer to solvefor x,z^2=3^{10}+3^x is x=(ln(z^2-59049))/(ln(3))