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

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

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

inverse

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

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

Solution steps

y = 10 ln (40 t - 1200)

Replace

t

with

y

t = 10 ln (40 y - 1200)

Solve for

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

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

f (x) = (x - 1) 3

tan (x^{2})

- 1 (\frac{5 - x}{3 x - 4})

y = \frac{8 x - 5}{7 x + 1}

f (x) = 7 x + 3 - 15

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