What is the inverse of f(x)=(e^{arctan(1+ln(x))}-1) ?

The inverse of f(x)=(e^{arctan(1+ln(x))}-1) is e^{tan(ln(x+1))-1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)=(e^{arctan(1+ln(x))}-1)

inverse

f (x) = (e^{a r c t a n (1 + l n (x))} - 1)

e^{t a n (l n (x + 1)) - 1}

Solution steps

Solve for

arctan (1 + ln (x)), y = e^{a r c t a n (1 + l n (x))} - 1

arctan (1 + ln (x)) = ln (y + 1)

arctan (a) = b \Rightarrow a = tan (b)

1 + ln (x) = tan (ln (y + 1))

Solve for

x, 1 + ln (x) = tan (ln (y + 1))

e^{t a n (l n (y + 1)) - 1}

Substitute

y = x

e^{t a n (l n (x + 1)) - 1}

1 - x

f (x) = 3^{(x - 2)}

f (x) = \frac{3 x + 4}{7}

f (x) = \frac{9 x}{4}

f (x) = arccos (0)

What is the inverse of f(x)=(e^{arctan(1+ln(x))}-1) ?
The inverse of f(x)=(e^{arctan(1+ln(x))}-1) is e^{tan(ln(x+1))-1}