What is the inverse of 25(1/5)^{x-2}-1 ?

The inverse of 25(1/5)^{x-2}-1 is -(ln(\frac{x+1)/(25))}{ln(5)}+2

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inverse of 25(1/5)^{x-2}-1

inverse

25 (\frac{1}{5})^{x - 2} - 1

- \frac{ln ( \frac{x + 1}{25} )}{ln ( 5 )} + 2

Solution steps

y = 25 (\frac{1}{5})^{x - 2} - 1

Replace

x

with

y

x = 25 (\frac{1}{5})^{y - 2} - 1

Solve for

y, x = 25 (\frac{1}{5})^{y - 2} - 1

- \frac{ln ( \frac{x + 1}{25} )}{ln ( 5 )} + 2

(4 - x) - (x^{2} - 3 x)

f (x) = \frac{1}{x + 19}

\frac{x}{x ^{2} + 3}

(- 4, - 5), (2, 1)

f (x) = \frac{x + 1}{x}

What is the inverse of 25(1/5)^{x-2}-1 ?
The inverse of 25(1/5)^{x-2}-1 is -(ln(\frac{x+1)/(25))}{ln(5)}+2