How do I solve log_3(2x-1)≤2 for x?

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Answer 1 · 2018-05-09T03:47:37

$\frac{1}{2} < x \leq 5$ $1/2 < x ≤ 5$ will be our solution

You're going to want to solve for $x$ $x$ .

$2 x - 1 \leq 3^{2}$ $2x - 1 ≤ 3^2$

$2 x - 1 \leq 9$ $2x - 1 ≤ 9$

$2 x \leq 10$ $2x ≤ 10$

$x \leq 5$ $x ≤ 5$

However, we need $2 x - 1 > 0$ $2x - 1 > 0$ , or $x > \frac{1}{2}$ $x > 1/2$ . Therefore, our answer is

$\frac{1}{2} < x \leq 5$ $1/2 < x ≤ 5$

We may confirm graphically.

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Hopefully this helps!