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 $1^{12}\neq 256$.) \[ \boxed{\text{Exactly one solution: }x=4} \]

Answer 2

Rather than expanding the cubic and spotting the pattern \((t-2)^3\), a shift of variable reveals a sum-of-cubes factorisation directly.

Setting up the shifted variable. Let \(L=\log_2x\) (so \(x=2^L\), and we need \(x>0\)). Then \(u=L^2-6L+12\), and the equation \(x^u=256=2^8\) becomes \(2^{Lu}=2^8\), i.e. \(Lu=8\), so \[ L(L^2-6L+12)=8. \] Let \(w=L-2\), so \(L=w+2\). Substituting: \[ L^2-6L+12=(w+2)^2-6(w+2)+12=w^2-2w+4. \] So the equation becomes \[ (w+2)(w^2-2w+4)=8. \]

Recognising the sum-of-cubes identity. Since \(a^3+b^3=(a+b)(a^2-ab+b^2)\), with \(a=w,\ b=2\): \[ (w+2)(w^2-2w+4)=w^3+2^3=w^3+8. \] So the equation is \(w^3+8=8\), giving \(w^3=0\), so \(w=0\) is the only real root (the other two cube roots of \(0\) all coincide at \(0\) as well, so there's no additional real solution). Hence \(L=2\), and \(x=2^2=4\).

1. Option "no solution": Ruled out — \(x=4\) genuinely satisfies the original equation, as can be checked directly: \(u=4-12+12=4\), and \(4^4=256\).
2. Option "exactly one solution": Matches — \(w^3=0\) has the single real root \(w=0\), giving the single value \(x=4\).
3. Option "exactly two distinct solutions": Ruled out — the cubic in \(w\) has a triple (not double) root at \(0\), so there is only one real value of \(x\), not two.
4. Option "exactly three distinct solutions": Ruled out — although \(w^3=0\) is a degree-3 equation, all three roots coincide at \(w=0\); there is no set of three distinct real \(x\)-values.

The shifted-variable factorisation confirms there is exactly one real \(x\) satisfying the equation, namely \(x=4\).

So the correct answer is exactly one solution for \(x\).