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: With \(L=\log_2x\), the equation reduces to \(L(L^2-6L+12)=8\). Substitute \(w=L-2\): this becomes \((w+2)(w^2-2w+4)=8\).

Step 2: By the sum-of-cubes identity, \((w+2)(w^2-2w+4)=w^3+8\), so the equation is \(w^3+8=8\), giving \(w^3=0\).

Step 3: The only real root is \(w=0\), so \(L=2\) and \(x=2^2=4\) — exactly one solution. \[ \boxed{\text{Exactly one solution: }x=4} \]