What is the answer to solvefor x,2^x+2^y=18 ?

The answer to solvefor x,2^x+2^y=18 is x=(ln(18-2^y))/(ln(2))

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solvefor x,2^x+2^y=18

solve for

x, 2^{x} + 2^{y} = 18

x = \frac{ln ( 18 - 2 ^{y} )}{ln ( 2 )}

Solution steps

2^{x} + 2^{y} = 18

Move

2^{y}

to the right side

2^{x} = 18 - 2^{y}

Apply exponent rules

x ln (2) = ln (18 - 2^{y})

Solve

x ln (2) = ln (18 - 2^{y}) : x = \frac{ln ( 18 - 2 ^{y} )}{ln ( 2 )}

x = \frac{ln ( 18 - 2 ^{y} )}{ln ( 2 )}

2^{n} = 60000

2^{5 x - 2} = 4^{x}

0. 2^{x + 1} = 125

7. 5^{3 r} = 2.6

4^{x} - 3 \cdot 2^{x} + 2 = 0

What is the answer to solvefor x,2^x+2^y=18 ?
The answer to solvefor x,2^x+2^y=18 is x=(ln(18-2^y))/(ln(2))