What is the inverse of y=(1-3^{2x})/(9^x-3) ?

The inverse of y=(1-3^{2x})/(9^x-3) is (ln(\frac{1+3x)/(x+1))}{2ln(3)}

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inverse of y=(1-3^{2x})/(9^x-3)

inverse

y = \frac{1 - 3 ^{2 x}}{9 ^{x} - 3}

\frac{ln ( \frac{1 + 3 x}{x + 1} )}{2 ln ( 3 )}

Solution steps

y = \frac{1 - 3 ^{2 x}}{9 ^{x} - 3}

Replace

x

with

y

x = \frac{1 - 3 ^{2 y}}{9 ^{y} - 3}

Solve for

y, x = \frac{1 - 3 ^{2 y}}{9 ^{y} - 3}

\frac{ln ( \frac{1 + 3 x}{x + 1} )}{2 ln ( 3 )}

f (x) = \frac{3 x}{8 x - 5}

f (x) = \frac{( 9 - x ^{2} )}{4}

(x + 4)^{4}

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

f (x) = \frac{4 x - 7}{x - 1}

What is the inverse of y=(1-3^{2x})/(9^x-3) ?
The inverse of y=(1-3^{2x})/(9^x-3) is (ln(\frac{1+3x)/(x+1))}{2ln(3)}