Exponential Functional Equation

Algebra Level 1

If $f(\ln x)=x^2+x+1$ , where $x>0$ , find $f(x)$ .

e^x+1

e^{2x}+e^x+1

e^x

\dfrac{1}{x}

2 solutions

Brian Charlesworth
Oct 9, 2014

Let $v = \ln(x)$ . Then $x = e^{v}$ , and the given equation becomes

$f(v) = (e^{v})^{2} + e^{v} + 1 = e^{2v} + e^{v} + 1$ .

Since $v$ is a dummy variable we can just rename it $x$ , giving us

$\boxed{f(x) = e^{2x} + e^{x} + 1}$ as the correct option.

Ahmad Ibrahim
Oct 10, 2014

$f(lnx)=x^{ 2 }+x+1\\ so\quad f(e^{ (lnx) })=(e^{ x })^{ 2 }+e^{ x }+1\\ giving\quad f(x)=e^{ 2x }+e^{ x }+1$