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?

2. Re: [ASK]Logarithmic Inequation 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?

5. Re: [ASK]Logarithmic 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 $(a-2)^2 \geq 0$

6. Re: [ASK]Logarithmic Inequation

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

7. Re: [ASK]Logarithmic Inequation

More detailed explanation

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

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