Find the value of k , so that the function $f (x)$ is continuous at the indicated point
$f (x) \begin{matrix} = \frac{8^{x} - 2^{x}}{k^{x} - 1} f o r x \neq o = 2 x = 0 \end{matrix}} a t x = 0$

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Solution

$\begin{matrix} lt x \to 0 \frac{8^{x} - 2^{x}}{k^{x} - 1} = 2 lt x \to 0 \frac{8^{x} ln 8 - 2^{x} ln 2}{k^{x} ln k} = 2 x = 0 \frac{ln 8 - ln 2}{ln k} = 2 ln 2 = 2 ln k k^{2} = 2 k = \pm \sqrt{2} \end{matrix}$

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