[ASK]Logarithmic Inequation

Apr 2018
182
38
Sukoharjo
How to solve this?
\(\displaystyle (log_2x)^2+4>5log_7x+log_3x^2\)
This is what I've done so far (attached):

How to solve the quadratic inequation formed? Factorization is an obvious no, and the quadratic formula will just make everything messier, is there any other way?
 

Attachments

Plato

MHF Helper
Aug 2006
22,492
8,653
How to solve this?
\(\displaystyle (log_2x)^2+4>5log_7x+log_3x^2\)
This is what I've done so far (attached):
How to solve the quadratic inequation formed? Factorization is an obvious no, and the quadratic formula will just make everything messier, is there any other way?
I would change to the standard base: $\log_2(x)=\dfrac{\log(x)}{\log(2)},~\log_7(x)$$= \dfrac{\log(x)}{\log(7)},~\&~\log_3(x^2)= \dfrac{\log(x^2)}{\log(3)}$
 
Jun 2013
1,134
605
Lebanon
Assume $x>1$

Let
\(\displaystyle a=\log _2x\)

\(\displaystyle b=\log _3x\)

\(\displaystyle c=\log _7x\)

\(\displaystyle 2^a=3^b=7^c\)

Show $b<a$ and $5c<2a$

and therefore

\(\displaystyle 5c+2b<4a\leq a^2+4\)
 
Apr 2018
182
38
Sukoharjo
I would change to the standard base: $\log_2(x)=\dfrac{\log(x)}{\log(2)},~\log_7(x)$$= \dfrac{\log(x)}{\log(7)},~\&~\log_3(x^2)= \dfrac{\log(x^2)}{\log(3)}$
I did that before submitting the one I posted in the attachment, and it didn't help much if at all.

Show $b<a$ and $5c<2a$

and therefore

\(\displaystyle 5c+2b<4a\leq a^2+4\)
Where did that 5c < 2a come from?
And also, how did you get the last inequation?
 
Jun 2013
1,134
605
Lebanon
Where did that 5c < 2a come from?

it comes from $$2^5<7^2$$

And also, how did you get the last inequation?[/QUOTE]

$4a\leq a^2+4$ comes from $(a-2)^2 \geq 0$
 
Apr 2018
182
38
Sukoharjo
I think I am having a hard time comprehending this...
 
Jun 2013
1,134
605
Lebanon
More detailed explanation

$$2^5<7^2$$
$$2^{5c}<7^{2c}=2^{2a}$$
$$5c<2a$$
$$2b<2a$$

Add

$$5c+2b<4a\leq a^2+4$$
 
Apr 2018
182
38
Sukoharjo
Okay, I understand the 5c + 2b < 4a part now, but why do we need that a² + 4? You said that it came from (a – 2)², but where did we get that from?