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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. $\log_{2}27$

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.

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

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

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

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

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