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

We need a strategy to simplify this exponential equation. The key insight is to notice that both terms involve powers of 2 with multiples of $3x$.

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Let's rewrite $2^{3x+2}$ using the rule $a^{m+n} = a^m \cdot a^n$:

$2^{3x+2} = 2^{3x} \cdot 2^2 = 2^{3x} \cdot 4$

Also, $2^{6x} = (2^{3x})^2$ using the rule $(a^m)^n = a^{mn}$

So our equation becomes:

$(2^{3x})^2 + 4 \cdot 2^{3x} - 21 = 0$

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When we see the same expression appearing multiple times (like $2^{3x}$), substitution helps us convert a complex exponential equation into a simple quadratic equation.

Let $y = 2^{3x}$

Then $(2^{3x})^2 = y^2$

Substituting into our equation:

$y^2 + 4y - 21 = 0$

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We can factor this quadratic. We need two numbers that multiply to $-21$ and add to $4$.

Those numbers are $7$ and $-3$ because: $7 \times (-3) = -21$ and $7 + (-3) = 4$

$(y + 7)(y - 3) = 0$

Therefore: $y = -7$ or $y = 3$

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Since $y = 2^{3x}$ and any positive number raised to any real power is always positive, we must have $y > 0$.

$y = -7$ is invalid because $2^{3x}$ can never be negative

$y = 3$ is valid because it's positive

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Since $y = 3$:

$2^{3x} = 3$

To solve for $x$, we take the logarithm base 2 of both sides:

$3x = \log_2 3$

Therefore: $x = \dfrac{\log_2 3}{3}$

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Answer: $x = \dfrac{\log_2 3}{3}$

When solving exponential equations, look for opportunities to use substitution by identifying repeated exponential expressions. This transforms the problem into a familiar algebraic equation that's much easier to solve.