Question

Let $u=(\log_2 x)^2-6\log_2 x+12$ where $x$ is a real number. Then the equation $x^u=256$ has:

• A. no solution for $x$
• B. exactly one solution for $x$
• C. exactly two distinct solutions for $x$
• D. exactly three distinct solutions for $x$

Hint:

When an exponent depends on $\log$ of the base, set $t=\log_b x$ so that $x=b^t$ and rewrite everything in base $b$. Often the resulting polynomial factorizes neatly.

Answer 1

The Correct Option is B

Step 1: Domain and substitution.
We need $x > 0$ (so that $\log_2 x$ is defined and $x^u$ makes sense). Let \[ t=\log_2 x \quad\Rightarrow\quad x=2^t,\qquad u=t^2-6t+12=(t-3)^2+3. \] Step 2: Convert the equation to base 2.
\[ x^u=(2^t)^{\,u}=2^{\,tu}=256=2^8 \quad\Rightarrow\quad tu=8. \] Hence \[ t\,(t^2-6t+12)=8 \ \Rightarrow\ t^3-6t^2+12t-8=0. \] Step 3: Factor the cubic.
Notice \[ (t-2)^3=t^3-6t^2+12t-8, \] so \[ (t-2)^3=0 \ \Rightarrow\ t=2 \ (\text{triple root}). \] Step 4: Back-substitute for $x$.
\[ t=\log_2 x=2 \ \Rightarrow\ x=2^2=4. \] This yields a single real $x$. (Note $x=1$ would give $u=12$ but $1^{12}\neq 256$.) \[ \boxed{\text{Exactly one solution: }x=4} \]