What is the answer to solvefor x,(2007)^x+(2007)^y=1 ?

The answer to solvefor x,(2007)^x+(2007)^y=1 is x=(ln(1-2007^y))/(ln(2007))

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solvefor x,(2007)^x+(2007)^y=1

solve for

x, (2007)^{x} + (2007)^{y} = 1

x = \frac{ln ( 1 - 200 7 ^{y} )}{ln ( 2007 )}

Solution steps

(2007)^{x} + (2007)^{y} = 1

Move

200 7^{y}

to the right side

200 7^{x} = 1 - 200 7^{y}

Apply exponent rules

x ln (2007) = ln (1 - 200 7^{y})

Solve

x ln (2007) = ln (1 - 200 7^{y}) : x = \frac{ln ( 1 - 200 7 ^{y} )}{ln ( 2007 )}

x = \frac{ln ( 1 - 200 7 ^{y} )}{ln ( 2007 )}

n = \frac{( 2 ^{n + 4} - 2 ^{n + 3} )}{( 2 ^{n + 4} )}

\frac{8 1 ^{2}}{( 72 9 ^{1} ) ^{- x}} = (9^{2})^{x}

2^{x - 4} = 4. 2^{x - 6}

2 = 1.0 6^{x}

x - e^{- x} = 0

What is the answer to solvefor x,(2007)^x+(2007)^y=1 ?
The answer to solvefor x,(2007)^x+(2007)^y=1 is x=(ln(1-2007^y))/(ln(2007))