Solve.
-$(\log 3 x)^{2}+2 \log 3 x-8=0$
A) $\{9\}$
B) $\left\{\frac{1}{81}\right\}$
C) $\left\{\frac{1}{81}, 9\right\}$
D) $\{4\}$

Question

Quizplus · Accepted Answer

This equation can be solved by treating $\log 3 x$ as a single variable, say $y$, turning the equation into a quadratic equation: $y^2 + 2y - 8 = 0$. Factoring or using the quadratic formula gives $y = 2$ or $y = -4$. Substituting back for $y = \log 3 x$, we get $\log 3 x = 2$ and $\log 3 x = -4$. Solving each equation for $x$ gives $x = 9$ and $x = \frac{1}{81}$, respectively. Therefore, both $x = 9$ and $x = \frac{1}{81}$ are solutions.

Solve - (log⁡3x)2+2log⁡3x−8=0(\log 3 x)^{2}+2 \log 3 x-8=0(log3x)2+2log3x−8=0 A) {9}\{9\}{9} B) {181}\left\{\frac{1}{81}\right\}{811​} C) {181,9}\left\{\frac{1}{81}, 9\right\}{811​,9}

Solve
- $(\log 3 x)^{2}+2 \log 3 x-8=0$ A) $\{9\}$ B) $\left\{\frac{1}{81}\right\}$ C) $\left\{\frac{1}{81}, 9\right\}$