What is the answer to solvefor x,2^x=((1+y))/((1-y)) ?

The answer to solvefor x,2^x=((1+y))/((1-y)) is x=(ln(\frac{1+y)/(1-y))}{ln(2)}

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solvefor x,2^x=((1+y))/((1-y))

solve for

x, 2^{x} = \frac{( 1 + y )}{( 1 - y )}

x = \frac{ln ( \frac{1 + y}{1 - y} )}{ln ( 2 )}

Solution steps

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

Apply exponent rules

x ln (2) = ln (\frac{1 + y}{1 - y})

Solve

x ln (2) = ln (\frac{1 + y}{1 - y}) : x = \frac{ln ( \frac{1 + y}{1 - y} )}{ln ( 2 )}

x = \frac{ln ( \frac{1 + y}{1 - y} )}{ln ( 2 )}

\frac{( 5 ^{5} \cdot 5 ^{- 4} \cdot 5 ^{x} )}{( 5 ^{12} )} = 5^{- 5}

2^{2 x} = 5^{x - 1}

b, (a^{b})^{c} = 512

(1) 2^{2 x - 1} = \frac{1}{( 8 ^{x - 3} )}

x, 5 \cdot 1 f \cdot 5^{2 x - 1} - 2 5^{x - 1} = 200

What is the answer to solvefor x,2^x=((1+y))/((1-y)) ?
The answer to solvefor x,2^x=((1+y))/((1-y)) is x=(ln(\frac{1+y)/(1-y))}{ln(2)}