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If $f(x)=\frac{(81)^{x}-(9)^{x}}{(k)^{x}-1}$ if $x \neq 0$
$=2 \quad$ if $\quad x=0$
is continuous at $x=0$, then the value of $k$ is
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$=2 \quad$ if $\quad x=0$
is continuous at $x=0$, then the value of $k$ is
Solution:
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Verified Answer
The correct answer is:
3
$\lim _{x \rightarrow 0} \frac{81^{x}-9^{x}}{k^{x}-1}=2$
$\therefore \lim _{x \rightarrow 0} \frac{9^{x}\left(9^{x}-1\right)}{k^{x}-1}=2 \Rightarrow \frac{\left(\lim _{x \rightarrow 0} 9^{x}\right)\left(\lim _{x \rightarrow 0} \frac{9^{x}-1}{x}\right)}{\lim _{x \rightarrow 0} \frac{k^{x}-1}{x}}=2$
$\frac{\log 9}{\log k}=2 \Rightarrow \log 9=2 \log k \Rightarrow \log k^{2}=\log 9 \Rightarrow k^{2}=9 \Rightarrow k=3$
$\therefore \lim _{x \rightarrow 0} \frac{9^{x}\left(9^{x}-1\right)}{k^{x}-1}=2 \Rightarrow \frac{\left(\lim _{x \rightarrow 0} 9^{x}\right)\left(\lim _{x \rightarrow 0} \frac{9^{x}-1}{x}\right)}{\lim _{x \rightarrow 0} \frac{k^{x}-1}{x}}=2$
$\frac{\log 9}{\log k}=2 \Rightarrow \log 9=2 \log k \Rightarrow \log k^{2}=\log 9 \Rightarrow k^{2}=9 \Rightarrow k=3$
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