Given the equation: (3 x)2 - 5 3 x + 6 = 0.

Question

ScanSolve AI Tutor · Accepted Answer

Here's the solution to the problem:

Given the equation: $(\log_3 x)^2 - 5 \log_3 x + 6 = 0$ .

Let $y = \log_3 x$ .
Substitute $y$ into the given equation to form a quadratic equation.
$y^2 - 5y + 6 = 0$
Factor the quadratic equation.
We need two numbers that multiply to 6 and add to -5. These numbers are -2 and -3.
$(y - 2)(y - 3) = 0$
Solve for the possible values of $y$ .
$y - 2 = 0 \quad \text{or} \quad y - 3 = 0$ $y = 2 \quad \text{or} \quad y = 3$
Substitute back $\log_3 x$ for $y$ and solve for $x$ in each case.
Case 1: $y = 2$
$\log_3 x = 2$
Convert the logarithmic equation to an exponential equation: $x = 3^2$ .
$x = 9$
Case 2: $y = 3$
$\log_3 x = 3$
Convert the logarithmic equation to an exponential equation: $x = 3^3$ .
$x = 27$
The possible values of $x$ are $\boxed{9}$ and $\boxed{27}$ .

Answer