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

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To solve the quadratic inequality (log_2x)^2 + 4 > 5log_7x + log_3x^2, the discussion highlights the challenges of factorization and the quadratic formula, suggesting an alternative approach. The transformation of the inequality leads to a quadratic form 4y^2 - by + 1 > 0, where y = log_3(x) and b is approximately 3.04. The solution involves determining the intervals for y based on the properties of the quadratic function, specifically its upward-opening parabola. Clarifications are sought on the treatment of ln(2) and the correct expression under the square root, questioning whether it should be b^2 - 4 or b^2 - 16. The conversation emphasizes the need for careful manipulation of logarithmic expressions in solving the inequality.
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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?
 

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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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HallsofIvy said:
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>
 
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