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

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

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

solve for

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

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

Solution steps

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

Multiply both sides by

2^{y} + 1

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

Simplify

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

Expand

x (2^{y} + 1) : x \cdot 2^{y} + x

2^{y} - 1 = x \cdot 2^{y} + x

Move

1

to the right side

2^{y} = x \cdot 2^{y} + x + 1

Move

x 2^{y}

to the left side

2^{y} - x \cdot 2^{y} = x + 1

Factor

2^{y} - x \cdot 2^{y} : 2^{y} (1 - x)

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

Divide both sides by

1 - x

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

Apply exponent rules

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

3^{x} = 171

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

x, \frac{( 9 ^{x} - 1 )}{( 3 ^{y} + 2 )} = 81

(\frac{1}{5})^{3 + x} = (\frac{1}{25})^{4}

1 1^{- x - 5} = 1 4^{3 x}

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