Question

If $f\left(x\right)=2^x$ then $\frac{f(x+3)}{f(x-1)}=$

• A. $f\left(x\right)$
• B. $\frac{1}{f\left(x\right)}$
• C. $f\left(4\right)$
• D. $f\left(2\right)$

Answer 1

The correct option is C $f\left(4\right)$

Let$f\left(x\right)=2^x$
$f(x+3)=2^{x+3}=2^x\times 2^3$
$f(x-1)=2^{x-1}=\frac{2^x}{2}[\because a^{m+n}=a^m\times a^n]$
$\Rightarrow \frac{f(x+3)}{f(x-1)}=\frac{2^x\times 2^3}{\frac{2^x}{2}}=16=2^4=f\left(4\right)$