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?

Originally Posted by Monoxdifly
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)$$= \dfrac{\log(x)}{\log(7)},~\&~\log_3(x^2)= \dfrac{\log(x^2)}{\log(3)} 3. ## Re: [ASK]Logarithmic Inequation Assume x>1 Let \displaystyle a=\log _2x \displaystyle b=\log _3x \displaystyle c=\log _7x \displaystyle 2^a=3^b=7^c Show b<a and 5c<2a and therefore \displaystyle 5c+2b<4a\leq a^2+4 4. ## Re: [ASK]Logarithmic Inequation Originally Posted by Plato I would change to the standard base: \log_2(x)=\dfrac{\log(x)}{\log(2)},~\log_7(x)$$= \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.

Originally Posted by Idea
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?

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 $(a-2)^2 \geq 0$

I think I am having a hard time comprehending this...

More detailed explanation

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

$$5c+2b<4a\leq a^2+4$$