Question

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

• A. \(\frac{log_27}{3}\)
• B. \(log_2\ 9\)
• C. \(\frac{log_2\ 3}{3}\)
• D. \(log_2\ 27\)

Answer 1

The Correct Option is C

Given the equation: \[ 2^{6x} + 2^{3x} \cdot 2^2 - 21 = 0 \] 

Let us make a substitution: \[ \text{Let } 2^{3x} = y \] Then the equation becomes: \[ y^2 + 4y - 21 = 0 \]

Now solve this quadratic equation: \[ y^2 + 4y - 21 = 0 \Rightarrow (y - 3)(y + 7) = 0 \Rightarrow y = 3 \text{ or } y = -7 \]

But since \( y = 2^{3x} \), and exponential expressions are always positive: \[ 2^{3x} = 3 \text{ or } 2^{3x} = -7 \quad (\text{Not valid since } 2^{3x} > 0) \]

So only: \[ 2^{3x} = 3 \Rightarrow 3x = \log_2 3 \Rightarrow x = \frac{\log_2 3}{3} \]

Final Answer:

\[ \boxed{x = \frac{\log_2 3}{3}} \]