If x^((log₂x)² − 6log₂x + 1) = 64, Then x Is Equal To

A standard logarithmic equation problem, not tied to a specific exam paper.

Problem

If $x^{(\log_2x)^2-6\log_2x+1}=64$, then $x$ is equal to?

Key insight. Taking $\log_2$ of both sides pulls the exponent down as a multiplying factor — and substituting $t=\log_2x$ turns the whole equation into an ordinary cubic in $t$. Once factored, one factor is a sum of squares that can never be zero, leaving only one genuine solution.

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Approach

Take $\log_2$ of both sides, bringing the exponent down as a multiplying factor. Substitute $t=\log_2x$ to turn the resulting equation into a cubic in $t$. Factor the cubic, discard the factor that can never be zero, and solve for $t$, then convert back to $x$.

Solution

Step 1 — Take log base 2 of both sides

$$\left[(\log_2x)^2-6\log_2x+1\right]\cdot\log_2x = \log_264 = 6$$

Step 2 — Substitute t = log₂x

$$t(t^2-6t+1) = 6$$

$$t^3-6t^2+t-6=0$$

Step 3 — Factor the cubic

Testing $t=6$: $6^3-6(6)^2+6-6 = 216-216+6-6=0$ ✓ — so $(t-6)$ is a factor.

Dividing $t^3-6t^2+t-6$ by $(t-6)$:

$$t^3-6t^2+t-6 = (t-6)(t^2+1)$$

Step 4 — Solve the factored equation

$$(t-6)(t^2+1)=0$$

Since $t^2+1>0$ for every real $t$ (it can never equal zero), the only real solution comes from:

$$t-6=0 \implies t=6$$

Step 5 — Convert back to x

$$\log_2x = 6 \implies x=2^6=64$$

Answer

Common mistakes

• Trying to solve the cubic $t^3-6t^2+t-6=0$ by the quadratic formula or by expanding further. Testing small integer values (like $t=6$, suggested by the $64=2^6$ on the right-hand side) to find a factor first is far faster than any general cubic-solving method.
• Forgetting to check that $t^2+1$ has no real roots. Without this check, it’s easy to assume the cubic has three real solutions and continue searching for two more values of $x$ that don’t actually exist.

Practise next

• Solve $x^{(\log_3x)^2-4\log_3x+1}=81$ for $x$, using the same log-and-substitute approach.