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

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

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

inverse

\frac{4 ^{n - 1} + 1}{4 ^{1 - n} + 1}

\frac{ln ( 4 n )}{2 ln ( 2 )}

Solution steps

y = \frac{4 ^{n - 1} + 1}{4 ^{1 - n} + 1}

Replace

n

with

y

n = \frac{4 ^{y - 1} + 1}{4 ^{1 - y} + 1}

Solve for

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

\frac{ln ( 4 n )}{2 ln ( 2 )}

f (x) = x + 2 - 6

f (x) = x + 2 + 4

f (x) = lo g_{10} (x + 3) - 2

f (x) = 9 x^{5} - 7

f (x) = arctan (0)

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