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The real root of the equation $2^{6x}+2^{3x+2}-21=0$ is

  1. $\frac{\log_{2}7}{3}$
  2. $\log_{2}9$
  3. $\frac{\log_{2}3}{3}$
  4. $\log_{2}27$
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Given that, $2^{6x}+2^{3x+2}-21 =0$

$ \Rightarrow \left ( {2^{3x}} \right)^2 +2^{3x} \cdot 2^2-21 =0  \quad \longrightarrow (1)$

Let $2^{3x}$ be $k.$

Now, from equation $(1),$ we get

$ \Rightarrow k^2 +4k-21 =0$

$ \Rightarrow k^2 +7k -3k-21 =0$

$ \Rightarrow k(k+7)-3(k+7)=0$

$ \Rightarrow (k+7)(k-3)=0$

$ \Rightarrow \boxed {k=-7,3}$

$ \Rightarrow k \neq {-7}$, because $2^{3x}$ not be $-7$ for any value of $x.$

So, $ \boxed {k =3}$

$ \Rightarrow 2^{3x}=3$

Taking $\log_{2}$on the both sides.

$ \Rightarrow\log_{2}2^{3x} =\log_{2}3$

$ \Rightarrow 3x \log_{2}2 = \log_{2}3$

$ \Rightarrow 3x = \log_{2}3 \quad (\because \log_{a}a = 1)$

$ \Rightarrow \boxed{ x = \frac {\log_{2}3}{3}}$

Correct Answer: C
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