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

Monoxdifly

Well-known member
Aug 6, 2015
271
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?
 

Attachments

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Your use of a superscript-prefix for logarithm base, [tex]^3log x[/tex], rather than a subsript, [tex]log_3(x)[/tex], is a bit confusing and sometimes very misleading as what should be [tex]2 log_3(x)[/tex] is given as [tex]2^3 log(x)[/tex] 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 [tex]4(log_x(2))^2- (5 log_7(2)+ 2log_3(2)) log_x(2)+ 1> 0[/tex]. I would write that [tex]4y^2- by+ 1> 0[/tex] with [tex]y= log_3(x)[/tex] and b= 5 log7(2)+ 2log3(2) which is approximately 3.04.

By the quadratic formula, y is 0 for [tex]\frac{-b\pm\sqrt{b^2- 4}}{8}[/tex]. 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 [tex]y= log_x(2)= \frac{ln(x)}{ln(2)}> \frac{-b+\sqrt{b^2- 4}}{8}[/tex] or [tex]y= log_x(2)= \frac{ln(x)}{ln(2)}< \frac{-b-\sqrt{b^2- 4}}{8}[/tex]. Since ln(2) is positive, that is the same as [tex]ln(x)>
\frac{-b+\sqrt{b^2- 4}}{8}
[/tex] so [tex]x> e^{\frac{-b+\sqrt{b^2- 4}}{8}}[/tex]
and [tex]x< e^{ln(2)\frac{-b-\sqrt{b^2- 4}}{8}}[/tex]
. Use b= 3.04, approximately, to get a numeric answer if you need it.
 
Last edited:

Monoxdifly

Well-known member
Aug 6, 2015
271
Since ln(2) is positive, that is the same as [tex]ln(x)>
\frac{-b+\sqrt{b^2- 4}}{8}
How did the ln(2) vanish just because it's positive? Also, are you sure that what's in the square roots should be \(\displaystyle b^2-4\) instead of \(\displaystyle b^2-16\)?