MHB [ASK]Quadratic Inequation (log_2x)^2+4>5log_7x+log_3x^2

[Monoxdifly]

How to solve this?
$$(log_2x)^2+4>5log_7x+log_3x^2$$
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?

 

[HallsofIvy]

Your use of a superscript-prefix for logarithm base, ^3log x, rather than a subsript, log_3(x), is a bit confusing and sometimes very misleading as what should be 2 log_3(x) is given as 2^3 log(x) where the base of the logarithm looks like an exponent on 2. In any case, if I am reading it correctly you wind up with 4(log_x(2))^2- (5 log_7(2)+ 2log_3(2)) log_x(2)+ 1> 0. I would write that 4y^2- by+ 1> 0 with y= log_3(x) and b= 5 log₇(2)+ 2log₃(2) which is approximately 3.04.

By the quadratic formula, y is 0 for \frac{-b\pm\sqrt{b^2- 4}}{8}. Since the leading coefficient of the quadratic is positive, the graph of the quadratic is a parabola opening upward. The inequality will be satisfied for y= log_x(2)= \frac{ln(x)}{ln(2)}&gt; \frac{-b+\sqrt{b^2- 4}}{8} or y= log_x(2)= \frac{ln(x)}{ln(2)}&lt; \frac{-b-\sqrt{b^2- 4}}{8}. Since ln(2) is positive, that is the same as ln(x)&gt; <div style="text-align: left"><span style="font-family: 'Verdana'">\frac{-b+\sqrt{b^2- 4}}{8}</span>&#8203;</div><span style="font-family: 'Verdana'"><br /> </span> so x&gt; e^{<span style="font-family: 'Verdana'"><span style="font-family: 'Verdana'">\frac{-b+\sqrt{b^2- 4}}{8}}</span></span>

and x&lt; e^{ln(2)<span style="font-family: 'Verdana'">\frac{-b-\sqrt{b^2- 4}}{8}}</span>

. Use b= 3.04, approximately, to get a numeric answer if you need it.​

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[Monoxdifly]

Since ln(2) is positive, that is the same as ln(x)&gt; <div style="text-align: left"><span style="font-family: 'Verdana'">\frac{-b+\sqrt{b^2- 4}}{8}</span>&#8203;</div><span style="font-family: 'Verdana'"><br /> </span>

<span style="font-family: 'Verdana'"><br /> <br /> How did the ln(2) vanish just because it&#039;s positive? Also, are you sure that what&#039;s in the square roots should be $$b^2-4$$ instead of $$b^2-16$$?</span>