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I have this problem that I can solve halfway, but I'm struggling to find the interval for the solution.The inequality is this: $$\log_3(4^x+1)+\log_{4^x+1}(3)>2.5.$$

Now here is the method how I tried to solve this inequality:

$\log_3(4^x+1)+ \frac{1}{\log_{3}(4^x+1)}>2.5$

Substitute $\log_3(4^x+1)$ with $u$:

$\log_3(4^x+1)=u$

$u+ \frac{1}{u}>2.5$ multiply both side with $u$

$u^2-2.5u+1>0$

Now solving for $u$ I get $u_1=1$ and $u_2=2$

Next: going back to the substitution:

$\log _3(4^x+1)=u$

How do I proceed from now on assuming my calculations are right? How do I find the intervals?

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We need to solve $$u^2-2.5u+1>0$$ or $$(u-2)(u-0.5)>0,$$

which gives $$u>2$$ or $$u<\frac{1}{2}.$$ For $u>2$ we obtain $$\log_3(4^x+1)>2$$ or $$4^x>8$$ or $$x>\frac{3}{2}.$$ While for $u<\frac{1}{2}$ we obtain $$4^x<\sqrt3-1$$ or $$x<\log_4(\sqrt3-1)$$

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  • $\begingroup$ After finding the x(plural) do you then find the intersection or the union, what is the logic behind these two. How should one proceed? $\endgroup$ – L.B Aug 29 '17 at 19:20
  • $\begingroup$ @L.B I fixed my post. I hope now it's clear. $\endgroup$ – Michael Rozenberg Aug 29 '17 at 19:22
  • $\begingroup$ Could you please shed some light on the question I asked, I'm having much trouble figuring out if I should find the intersection or find the union. $\endgroup$ – L.B Aug 29 '17 at 19:25
  • $\begingroup$ @L.B You need to solve an inequality $u^2-2.5u+1>0$ that you are not doing. See please my solution. $\endgroup$ – Michael Rozenberg Aug 29 '17 at 19:30
  • $\begingroup$ I just wrote the solutions without pointing out the intervals. In the post I have written for u1 we have x1=1/2 etc. The inequality is $(0,1/2) v (2,+infinity)$. I don't know why sometimes it's finding the union and why sometimes( referring to different problems obviously) is finding the intersection, that's what is bothering me. If you can, please provide me an answer. Thank you. $\endgroup$ – L.B Aug 29 '17 at 19:37
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$log_3(4^x+1)+log_{4^x+1}(3)>2.5$

$\frac{ln(4^x+1)}{ln(3)}+\frac{ln(3)}{ln(4^x+1)}>2.5$

Before I go any further, let me establish something. Assuming x is finite a real number, $4^x$>0. This means $4^x+1$>1. ln($4^x$+1)>0 if $4^x$+1 is greater than 1, which I have just established it is. Now, continuing from where we left off:

$\frac{ln(4^x+1)}{ln(3)}+\frac{ln(3)}{ln(4^x+1)}>2.5$

Multiply both sides by $ln(4^x+1)$

$\frac{ln²(4^x+1)}{ln(3)}+ln(3)>2.5ln(4^x+1)$

Since $ln(4^x+1)>0$, we did not have to change the > sign.

$\frac{ln²(4^x+1)}{ln(3)}-2.5ln(4^x+1)+ln(3)>0$

Let's assign u to be ln(4$^x$+1)

$\frac{u²}{ln(3)}-\frac{5u}2+ln(3)>0$

If you graph this quadratic function out, you'll see the value is greater than zero when u is less than the first quadratic root or greater than the second. So, to finish up this inequality, let's solve for the roots of this quadratic:

$\frac{u²}{ln(3)}-\frac{5u}2+ln(3)=0$

$\frac{-B±\sqrt{B²-4AC}}{2A}$

A=$\frac{1}{ln(3)}$ B=$-\frac{5}2 $ C=ln(3)

$\frac{-(-\frac{5}2)±\sqrt{(-\frac{5}2)²-4(\frac{1}{ln(3)})(ln(3))}}{2\frac{1}{ln(3)}}$

$\frac{\frac{5}2±\sqrt{\frac{25}4-4}}{2\frac{1}{ln(3)}}$

$\frac{ln(3)(\frac{5}2±\sqrt{\frac{25}4-\frac{16}4})}2$

$\frac{ln(3)(\frac{5}2±\sqrt{\frac{9}4})}2$

$\frac{ln(3)(5±3)}4$

$\frac{ln(3)(8)}4$ or $\frac{ln(3)(2)}4$

2ln(3) or $\frac{ln(3)}2$

since ln(3)>0, the greater of the two would be 2ln(3).

u<$\frac{ln(3)}2$ or u>2ln(3)

As mentioned before, u=ln($4^x$+1)

$ln(4^x+1)<\frac{ln(3)}2 or ln(4^x+1)>2ln(3)$

$ln(4^x+1)<ln(\sqrt{3}) or ln(4^x+1)>ln(9)$

As e$^x$ is an increasing function over all values x is real, we can put e to the power of both sides of both equations.

$4^x+1<\sqrt{3} or 4^x+1>9$

$4^x<\sqrt{3}-1 or 4^x>8$

As log$_4$(x) is also increasing over all x is real, we can apply the function $log_4(x)$ over both sides of both equations.

$x<log_4(\sqrt{3}-1) or x>log_4(8)$

$x<log_4(\sqrt{3}-1) or x>\frac{3}2$

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After substituting $$log_{3}(4^{x}+1)=u$$the quadratic inequality $$u^{2}-2.5u+1>0$$ is obtained. Which can be solved by using the Wavy curve method. We get u< 0.5 or u>2. And on Substituting back, we get $$log_{3}(4^{x}+1)<0.5\Rightarrow x<log_{4}(3^{0.5}-1)$$ and $$log_{3}(4^{x}+1)>2 \Rightarrow x> \frac{3}{2} $$ Hence, The solution lies in the union of two intervals. That is, $$x\in (-∞,log_{4}(3^{0.5}-1)) \cup (\frac{3}{2},∞)$$

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  • $\begingroup$ @L.B One way to explain that is how quadratic functions(you quadratic equation in u) are on a graph(a parabola). if you a say the value of the function is greater than 0, it means that all the points where the graph is above x axis, which can be both, less than the first root or greater than the second root. Check this graph for clear understanding of my point $\endgroup$ – Urav Aug 29 '17 at 20:02
  • $\begingroup$ I think there is a logical error in the proof of the answer (the final answer is correct). The solution of the quadratic inequality (in $u$) is $u < 0.5$ OR $u > 2$. So the final solution is $x < \log_4(3^{0.5} - 1)$ OR $x > 3/2$, whence the union of the two intervals, not the intersection. $\endgroup$ – Taroccoesbrocco Aug 29 '17 at 20:07
  • $\begingroup$ Thank you. As mentioned by @Michael Rozenberg I wasn't thinking much of the quadratic. $\endgroup$ – L.B Aug 29 '17 at 20:07
  • $\begingroup$ @Taroccoesbrocco Yes, Rectified. Thankyou $\endgroup$ – Urav Aug 29 '17 at 20:11

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