Solving inequality of $ \log_{10}{(2x-5)} < \log_{1\over10}{(x-3)} $

Question

I have a problem to solving this inequality $$ \log_{10}{(2x-5)} < \log_{1\over10}{(x-3)} $$ My work so far is

Step 1 : Finding the definition term of $\log_{10}{(2x-5)}$ $$2x-5 > 0$$ $$2x>5$$ $$x>\frac{5}{2}$$

Step 2: Finding the definition term of $\log_{1\over10}{(x-3)}$ $$x-3>0$$ $$x>3$$

Step 3 : Finding the inequality term $$\log_{10}{(2x-5)} < \log_{1\over10}{(x-3)}$$ $$\log_{10}{(2x-5)} < \frac{1}{\log_{10}{(x-3)}} $$

From this position, i am confused, what should i do for the next step? Thank you for your help, sir.

After reading all suggestion, i try to redo my work also i realize my mistake in step 3.

Step 3 : $$ \log_{10}{(2x-5)} < \log_{1\over10}{x-3} $$ $$ \log_{10}{(2x-5)} < \log_{10}{(\frac{1}{x-3})} $$ $$ (2x-5) < \frac{1}{(x-3)}$$ $$ (2x-5)(x-3) < 1 $$ $$ 2x^2 -6x-5x+15 < 1 $$ $$ 2x^2-11x+15<1$$ $$2x^2-11x+15-1<0$$ $$2x^2-11x+14<0$$ $$(x-2)(2x-7)<0$$ $$2<x<\frac{7}{2}$$

Step 4: Finding the set of solutions in real number. Affect from definition term in step 1 to inequality term in step 3. $$ \frac{5}{2} < x< \frac{7}{2} $$ Affect from definition term in step 2 to inequality term in step 3. $$ 3 < x < \frac{7}{2} $$ $\therefore $ the set of solution is $\left(3<x<\frac{7}{2}, x \in R \right) $ Thanks for suggestion and explanation.

Answer 1

I will assume that all logarithms are defined (so $x>3$). Let $a=\log_{10} (2x-5)$ and $b=\log_{1/10} (x-3)$ . Then $10^{a}=2x-5$ and $10^{b}=\frac 1 {x-3}$. The inequality $a <b$ is equivalent to. $10^{a} <10^{b}$. So we can write the given inequality as $(2x-5)(x-3)<1$. Can you proceed from here?

The answer is $3<x<3.5$.

Answer 2

Be careful in your last step. According to the change of base formula:

$$\log_{1\over10}{(x-3)} = \frac{\log_{10} (x-3) }{\log_{10} (\frac{1}{10})} = -\log_{10} (x-3) \ne \frac{1}{\log_{10}{(x-3)}}$$

So you should have instead: $$\log_{10} (2x-5) < -\log_{10} (x-3)$$ $$10^{\log_{10} (2x-5)} < 10^{\log_{10} (x-3) \times-1}$$

Now simplify the right-hand side, and you should get a nice quadratic equation.

Answer 3

Steps 1 and 2 are good. However, there is a mistake on step 3. $\log_{1/10}(x-3)$ is not equal to $\frac{1}{\log_{10}(x-3)}$. You should go back to napierian logarithm with $$ \log_{1/10} (x-3) = \frac{\ln(x-3)}{\ln(1/10)} = \frac{\ln(x-3)}{- \ln(10)}$$ $$ \log_{10} (2x-5) = \frac{\ln(2x-5)}{\ln(10)}$$

Answer 4

First of all, we need $x>$max$(5/2,3)=3$

to keep $\log_{10}(2x-5),\log_{1/10}(x-3)$ real

$$\log_{10}(2x-5)<\log_{1/10}(x-3)=-\log_{10}(x-3)$$ as $\dfrac1{10}=10^{-1}$

$$\iff0>\log_{10}(2x-5)+\log_{10}(x-3)=\log_{10}(2x-5)(x-3)$$

$$\iff(2x-5)(x-3)<10^0=1$$

$$\iff2x^2-11x+14<0$$

$$\iff(2x-7)(x-2)<0$$

$$\implies2<x<\dfrac72$$

But we need to honor $x>3$

Can you take it from here?