Logarithmic inequality with substitution 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? 
 A: 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
$$x<\log_4(\sqrt3-1)$$
A: $log_3(4^x+1)+log_{4^x+1}(3)>2.5$
$\frac{ln(4^x+1)}{ln(3)}+\frac{ln(3)}{ln(4^x+1)}>2.5$
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.
$x<log_4(\sqrt{3}-1) or x>log_4(8)$
$x<log_4(\sqrt{3}-1) or x>\frac{3}2$
A: 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,
$$x\in (-∞,log_{4}(3^{0.5}-1)) \cup (\frac{3}{2},∞)$$
