What is the inverse of (4^x-2^{2x-1})/(4^x+1) ?

The inverse of (4^x-2^{2x-1})/(4^x+1) is (ln(-\frac{2x)/(2x-1))}{2ln(2)}

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inverse of (4^x-2^{2x-1})/(4^x+1)

inverse

\frac{4 ^{x} - 2 ^{2 x - 1}}{4 ^{x} + 1}

\frac{ln ( - \frac{2 x}{2 x - 1} )}{2 ln ( 2 )}

Solution steps

y = \frac{4 ^{x} - 2 ^{2 x - 1}}{4 ^{x} + 1}

Replace

x

with

y

x = \frac{4 ^{y} - 2 ^{2 y - 1}}{4 ^{y} + 1}

Solve for

y, x = \frac{4 ^{y} - 2 ^{2 y - 1}}{4 ^{y} + 1}

\frac{ln ( - \frac{2 x}{2 x - 1} )}{2 ln ( 2 )}

f (x) = - lo g_{2} (x + 1)

(1 + 2)^{n}

f (x) = x + 8 x

\frac{2 x + 1}{x + 1}

x (- x + 8)

What is the inverse of (4^x-2^{2x-1})/(4^x+1) ?
The inverse of (4^x-2^{2x-1})/(4^x+1) is (ln(-\frac{2x)/(2x-1))}{2ln(2)}