What is the inverse of y= 1/40 e^{0.1x}+30 ?

The inverse of y= 1/40 e^{0.1x}+30 is (ln(40x-1200))/(0.1)

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inverse of y= 1/40 e^{0.1x}+30

inverse

y = \frac{1}{40} e^{0.1 x} + 30

\frac{ln ( 40 x - 1200 )}{0.1}

Solution steps

y = \frac{1}{40} e^{0.1 x} + 30

Replace

x

with

y

x = \frac{1}{40} e^{0.1 y} + 30

Solve for

y, x = \frac{1}{40} e^{0.1 y} + 30

\frac{ln ( 40 x - 1200 )}{0.1}

f (x) = (x - 19)^{2}, [19, \infty]

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

9 sin (x)

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

\frac{1.05}{1 + 100 e ^{- 0.285 t}}

What is the inverse of y= 1/40 e^{0.1x}+30 ?
The inverse of y= 1/40 e^{0.1x}+30 is (ln(40x-1200))/(0.1)