Question

Find $k$, if $f$ is continuous at $x=0$ where
$f\left(x\right)=\{\begin{array}{c}\frac{{16}^x-2^x}{k^x-1}\text{when}x\ne 0\\ \\ 3\text{when}x=0\end{array}$

Answer 1

$f\left(x\right)=\frac{{16}^x-2^x}{k^x-1}$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{2^{4x}-2^x}{k^x-1}$

$=\underset{x\to 0}{lim}\frac{\frac{2^{4x}-2x}{x}}{\frac{k^x-1}{x}}$

$=\underset{x\to 0}{lim}\frac{\frac{2^{4x}-1-2^x+1}{x}}{\frac{k^x-1}{x}}$

$=\underset{x\to 0}{lim}\frac{(\frac{(2^{4x}-1)}{4x}\times 4)-\left(\frac{2^x-1}{x}\right)}{\frac{k^x-1}{x}}$

$=\frac{\left({\log }_{e}2\right)\times 4-\left({\log }_{e}2\right)}{{\log }_{e}k}$

$(\because \underset{x\to 0}{lim}\frac{a^x-1}{x}={\log }_{e}a)$

$=\frac{3{\log }_{e}2}{{\log }_{e}k}$

$=\frac{{\log }_e{2}^3}{{\log }_{e}k}$

For the function to be continuous

$\Rightarrow \frac{{\log }_e{2}^3}{{\log }_{e}k}=3$

$\Rightarrow {\log }_e{2}^3=3{\log }_{e}k$

$\Rightarrow {\log }_e{2}^3={\log }_e{k}^3$

$\therefore k=2$