Question

If $f\left(x\right)=e^x$ and $g\left(x\right)=\log e^x$, then which of the following is correct?

• A. $f\left(g\left(x\right)\right)\ne g\left(f\left(x\right)\right)$
• B. $f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)$
• C. $f\left(g\left(x\right)\right)+g\left(f\left(x\right)\right)=0$
• D. $f\left(g\left(x\right)\right)-g\left(f\left(x\right)\right)=1$

Answer 1

The correct option is B

$f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)$

Explanation for the correct option.

Find the correct relation between $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$.

For the functions $f\left(x\right)=e^x$ and $g\left(x\right)=\log e^x$, the function $f\left(g\left(x\right)\right)$ is defined as:

$\begin{array}{rcl}f\left(g\left(x\right)\right)& =& f\left(\log e^x\right)\\ & =& e^{\log e^x}\\ & =& e^x\left[b^{lo{g}_b\left(a\right)}=a\right]\end{array}$

Again the function $g\left(f\left(x\right)\right)$ is defined as:

$\begin{array}{rcl}g\left(f\left(x\right)\right)& =& g\left(e^x\right)\\ & =& \log e^{e^x}\\ & =& e^x\log e\left[log{a}^m=mloga\right]\\ & =& e^x\times 1\\ & =& e^x\end{array}$

Thus, it can be seen that $f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)$ and both are equal to $e^x$.

Hence, the correct option is B.