What is the inverse of f(x)=10ln(40x-1200) ?

The inverse of f(x)=10ln(40x-1200) is (e^{x/(10)}+1200)/(40)

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inverse of f(x)=10ln(40x-1200)

inverse

f (x) = 10 ln (40 x - 1200)

\frac{e ^{\frac{x}{10}} + 1200}{40}

Solution steps

y = 10 ln (40 x - 1200)

Replace

x

with

y

x = 10 ln (40 y - 1200)

Solve for

y, x = 10 ln (40 y - 1200)

\frac{e ^{\frac{x}{10}} + 1200}{40}

g (62) = \frac{100}{x - 12} + 8

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

ln (ln (ln (x)))

\frac{x + 4}{x}

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

What is the inverse of f(x)=10ln(40x-1200) ?
The inverse of f(x)=10ln(40x-1200) is (e^{x/(10)}+1200)/(40)