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

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

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solvefor y,(1+x)^y= 1/((1-x))

solve for

y, (1 + x)^{y} = \frac{1}{( 1 - x )}

y = - \frac{ln ( 1 - x )}{ln ( 1 + x )}

Solution steps

(1 + x)^{y} = \frac{1}{( 1 - x )}

Apply exponent rules

y ln (1 + x) = ln (\frac{1}{1 - x})

Solve

y ln (1 + x) = ln (\frac{1}{1 - x}) : y = - \frac{ln ( 1 - x )}{ln ( 1 + x )}

y = - \frac{ln ( 1 - x )}{ln ( 1 + x )}

x, q^{x} + 1 = 3

(a) 1.2 \cdot 1 0^{6} = 12 \cdot 1 0^{x}, x

\frac{( 2 ^{n + 1} + 2 ^{n} )}{( 2 ^{n + 4} - 2 ^{n - 1} )} = 1

8^{x - 3} = 4^{x + 1}

(- 3)^{x} = - 243

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