Find the inverse of the function f(x) = e^(3x

Sunday, September 25, 2011

Find the inverse of the function f(x) = e^(3x - 5).

Any function f(x) and its inverse $\displaystyle{{f}}^{{-{{1}}}}{\left({x}\right)}$ follow the
relation: $\displaystyle{f{{\left({{f}}^{{-{{1}}}}{\left({x}\right)}\right)}}}={x}$ .

For the function $\displaystyle{f{{\left({x}\right)}}}={{e}}^{{{3}{x}}}$
- 5) $\displaystyle,\to{\det{{e}}}{r}\min{e}{t}{h}{e}\in{v}{e}{r}{s}{e},{w}{e}{u}{s}{e}{t}{h}{e}{r}{e}{l}{a}{t}{i}{o}{n}$ f(f^-1(x)) = x
.

$\displaystyle{f{{\left({{f}}^{{-{{1}}}}{\left({x}\right)}\right)}}}=$
x

$\displaystyle{{e}}^{{{3}\cdot{{f}}^{{-{{1}}}}{\left({x}\right)}-{5}}}={x}$

Now
take the logarithm to base e for both the sides.

$\displaystyle{\log}_{{e}}$
(e^(3*f^-1(x) - 5)) = log_e x

Use the relation $\displaystyle{{\log{{a}}}}^{{b}}=$
b*log a $\displaystyle{\quad\text{and}\quad}$ log_b b = 1

$\displaystyle{\left({3}\cdot{{f}}^{{-{{1}}}}{\left({x}\right)}-{5}\right)}\cdot{\log}_{{e}}{e}={\log}_{{e}}$
x

$\displaystyle{3}\cdot{{f}}^{{-{{1}}}}{\left({x}\right)}-{5}={\log}_{{e}}$
x

$\displaystyle{3}\cdot{{f}}^{{-{{1}}}}{\left({x}\right)}-{5}={\ln{}}$
x

$\displaystyle{3}\cdot{{f}}^{{-{{1}}}}{\left({x}\right)}={5}+{\ln{}}$
x

$\displaystyle{{f}}^{{-{{1}}}}{\left({x}\right)}={\left({5}+{\ln{}}\right.}$
x)/3

The required inverse function is $\displaystyle{{f}}^{{-{{1}}}}{\left({x}\right)}={\left({5}+{\ln{}}\right.}$
x)/3

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Sunday, September 25, 2011