#### 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?

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#### Plato

MHF Helper
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)$$= \dfrac{\log(x)}{\log(7)},~\&~\log_3(x^2)= \dfrac{\log(x^2)}{\log(3)} #### Idea Assume x>1 Let $$\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$$ #### Monoxdifly I would change to the standard base: \log_2(x)=\dfrac{\log(x)}{\log(2)},~\log_7(x)$$= \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?

#### Idea

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

#### Monoxdifly

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

#### Idea

More detailed explanation

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

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