$\log_3(2x-1) + \log_3(x-1) < 1$

Question

Become $\log_3(2x^2 -3x+1) < \log_3(3)$

At first i thought become $\log_3(2x^2 -3x+1) - \log_3(3) <0$

$\log_3\frac{2x^2-3x+1}{3}<0$

(EDIT : i make it $\frac{2x^2-3x+1}{3}<0$

$ 2x^2-3x+1 <0$ i guess its where i wrong?)

But it is $2x^2-3x+1<3$ why the first one wrong?

See https://math.stackexchange.com/questions/3406843/solving-inequality-of-log-102x-5-log-1-over10x-3/3406855#3406855 — lab bhattacharjee, Oct 24 '19 at 09:19
The last two equations you posted are equivalent, I don't understand what you mean is wrong? — kingW3, Oct 24 '19 at 09:53

score 1 · Answer 1 · answered Oct 24 '19 at 09:00

The first one is wrong because $\log_3 (3) = 1$. In addition, it should be $2x^2-3x+1$ instead of $2x^2-3x+1$.

That being said, the reason you want to put both sides into logarithms with the same base is because you can raise both sides to the power of $3$. If $\log_3 x = \log_3 y$, then $x = y$ which simplifies your equation into a quadratic.

Don't forget to check if the values of $x$ by substituting in the original equation.

Kavi Rama Murthy · Answer 2 · 2019-10-24T09:17:51.583

0

The question has been edited. My first sentence refers to the earlier version.

$(2x-1)(x-1)=2x^{2}-3x+1$. You have $2x^{2}-3x-1$ in stead of $2x^{2}-3x+1$.

Since $\log_3$ is an increasing function the inequality becomes $2x^{2}-3x+1 <3$. Or $2x^{2}-3x-2 <0$. This gives $-1/2<x<2$ as you can see by completing the square. But we need $x>1$ for the logarithms to be defined so the final answer is $1<x<2$.

edited Oct 24 '19 at 09:17

answered Oct 24 '19 at 08:59

Kavi Rama Murthy

311,013

$\log_3(2x-1) + \log_3(x-1) < 1$

2 Answers2