Logarithmic Inequality (fractional base)

In this video I highlighted a common mistake students make in solving this kind of logarithmic inequality

[johnroberts]

It's some time since an old dude like me undertook his Maths degree. I wish some of my lecturers had delivered such a fine, well paced and crystal clear lesson as this. You are gifted sir, many thanks ... it was great fun and educational of course. The carefully constricted board presentation was greatly appreciated.

[rabinbharadwaja]

Hello, I am a retired design engineer. I have forgotten many of the basics mathematic problems. I started watching you videos that re-educates me what I have learned more than 50 years ago. You know how to make math easier. Thanks

[daniorugbani]

Prime Newton, you are simply amazing teacher. I can watch your videos all day without getting tired. Please keep it up and continue to educate us on how to avoid making basic mistakes in solving math problems.

[glorrin]

One of the easy math concepts that is commonly misunderstood: the domain. So, by definition it is not an easy concept.

[Christian_Martel]

I’m always mixed when the base is between 0 and 1, so I didn’t hesitate to change the base of the log from 1/2 to 2.

Domain of x: -3/2 < x < 2/3


log1/2(2-3x) > log1/2(3+2x)
-log2(2-3x) > -log2(3+2x)
+ log2(2-3x) < +log2(3+2x)

2-3x < 3+2x
5x > -1
x > -1/5

Solution: -1/5 < x < 2/3

[mcrow]

A very useful problem that explores several tricky concepts. You explained it very well. Thank you.

[Samir-zbxk]

Very good explanation thank you sir...

I think its worth mentioning that we must swap the inequality anytime we apply a decreasing function on both sides of an inequality.

In this question we applied (1/2)^x on both sides, hence we swapped the inequality.

It can easily be shown that (1/2)^x is decreasing by its derivative, ln(1/2)⋅(1/2)^x

(1/2)^x is always positive and ln(1/2) is negative, hence the derivative is always negative, and the function is always decreasing

[josephparrish]

That was a beautiful explanation! So clear and so complete! I love watching you work!.

[slavinojunepri]

Instead of reasoning on a positive logarithmic base less than 1, we can change both sides of the inequation to base 2. This will in turn change the direction of the inequality since log2(1/2) =-1.

[davidcroft]

The reasoning with the one half power can be simplified knowing that logarithm of base *number less than one* is a decreasing function, therefore we have to change the direction of the inequality (like moltiplying by -1 or doing the reciprocal)

[parigupta]

I divided both sides by -1 and changed the inequality sign, which gives
log 2 (2-3x)< log 2 (3+2x)
Which implies (2-3x) < (3+2x)

Btw, I really love your videos ❤❤

[Abby-hisf]

The way I see log is to change it to base 10.
So log(2-3x) to base 1/2 will be =log(2-3x) / log (1/2)
Also log log(3+2x) to base 1/2 will be = log(3+2x) / log (1/2)
Log (1/2)= log2^-1 = -log 2
now multiplying both side by -log 2 will shift the inequality. So log(2-3x) will be less than log(3+2x)

[michaelkeffer]

Another way to solve this inequality is to add 3x to both sides and subtract 3 from both sides and then divide by 5. That would eliminate the need to divide both sides by a negative sign and flip the inequality sign. In other words 2 - 3x < 3 + 2x becomes -1 < 5x and then (-1/5) < x, which is the same as x > (-1/5).

[raghvendrasingh]

❤
similar problem
(0.1)^(x^2+6) > (0.1)^(5x)
since base lies between 0 and 1
x^2+6 < 5x
x^2 - 5x + 6 < 0
(x - 2) (x - 3) < 0
2 < x < 3

[Aadarshiitbhu]

Solve the JEE Advanced paper, An exam in India which is toughest exam of the country... It has very really good maths problems.. I will like to see your version of solutions...