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#### Monoxdifly

##### Well-known member
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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#### HallsofIvy

##### Well-known member
MHB Math Helper
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 $$y= log_x(2)= \frac{ln(x)}{ln(2)}> \frac{-b+\sqrt{b^2- 4}}{8}$$ or $$y= log_x(2)= \frac{ln(x)}{ln(2)}< \frac{-b-\sqrt{b^2- 4}}{8}$$. Since ln(2) is positive, that is the same as $$ln(x)> \frac{-b+\sqrt{b^2- 4}}{8}$$ so $$x> e^{\frac{-b+\sqrt{b^2- 4}}{8}}$$
and $$x< e^{ln(2)\frac{-b-\sqrt{b^2- 4}}{8}}$$
. Use b= 3.04, approximately, to get a numeric answer if you need it.

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#### Monoxdifly

##### Well-known member
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 $$\displaystyle b^2-16$$?