Question

If $x^{(\log_2x)^2-6\log_2x+11}=64$, then x is equal to

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (A), (B), (D)

2, 4, 8

Answer 2

To solve the equation $x^{(\log_2x)^2-6\log_2x+11}=64$, we follow these steps:

1. Take logarithm on both sides: Since the equation involves $\log_2x$, it is convenient to take $\log_2$ on both sides of the equation. $\log_2\left(x^{(\log_2x)^2-6\log_2x+11}\right) = \log_2(64)$

2. Apply logarithm properties: Use the property $\log_b(M^P) = P \cdot \log_b(M)$ on the left side and evaluate the right side. The exponent $(\log_2x)^2-6\log_2x+11$ comes down as a multiplier. Also, $64 = 2^6$, so $\log_2(64) = \log_2(2^6) = 6$. $((\log_2x)^2-6\log_2x+11) \cdot \log_2x = 6$

3. Substitute a variable: Let $y = \log_2x$. This transforms the equation into a polynomial equation in $y$. $(y^2-6y+11) \cdot y = 6$ $y^3 - 6y^2 + 11y = 6$

4. Rearrange into a standard polynomial form: $y^3 - 6y^2 + 11y - 6 = 0$

5. Solve the cubic equation for y: We can find integer roots by testing divisors of the constant term (-6), which are $\pm1, \pm2, \pm3, \pm6$.

   • For $y=1$: $1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$. So, $y=1$ is a root.
   • For $y=2$: $2^3 - 6(2)^2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0$. So, $y=2$ is a root.
   • For $y=3$: $3^3 - 6(3)^2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0$. So, $y=3$ is a root.

   Since we found three roots for a cubic equation, these are all the roots. The roots are $y=1, y=2, y=3$.

6. Substitute back to find x: Now, use $y = \log_2x$ to find the values of $x$.

   • If $y=1$: $\log_2x = 1 \implies x = 2^1 \implies x = 2$
   • If $y=2$: $\log_2x = 2 \implies x = 2^2 \implies x = 4$
   • If $y=3$: $\log_2x = 3 \implies x = 2^3 \implies x = 8$

All these values of $x$ (2, 4, 8) are positive, satisfying the domain requirement for logarithms ($x > 0$).