• July 11th 2009, 10:21 AM
Logic
Greetings,
I have the below problem, which goes relatively fine up to where I find the roots. I guess every two roots are to form a multitude. I get the two logarithm ones fine, x varies from the smaller log. to the larger one.
I am unsure what happens when I have no x4. The answer is that x ranges from minus infinity to the other root. However, I do not understand why.
• July 11th 2009, 03:37 PM
Soroban
Hello, Logic!

I don't see an inequality . . . Is it "greater than"?
Also, why are you looking for four roots?

Quote:

$8\sqrt{2^{2x-2} - 2^x + 1} \;\;{\color{red}>}\;\;2^{2x} - 2^{x+2} + 7$
Those initial steps were excellent! . . . $4\sqrt{2^{2x} - 4\!\cdot\!2^x + 4} \;>\;\left(2^{2x} - 4\!\cdot\!2^x + 4\right) + 3$

Here's my variation of the solution . . .

Note that: . $2^{2x} - 4\!\cdot\!2^x + 4 \:=\:\left(2^x - 2\right)^2$

So we have: . $4\sqrt{(2^x-2)^2} \;>\;(2^x-2)^2 + 3 \quad\Rightarrow\quad 4(2^x - 2) \;>\;(2^x-2)^2 + 3$

Let $t \:=\:2^x-2$

We have: . $4t \;>\;t^2 +3 \quad\Rightarrow\quad -t^2 + 4t - 3 \;>\;0$

Multiply by -1: . $t^2 - 4t + 3 \;<\;0 \quad\Rightarrow\quad (t-1)(t-3) \;<\;0$

And we have: . $\begin{array}{cccccc} t-1 \:>\:0 & \Rightarrow & t \:>\:1 \\ t-3 \:<\:0 &\Rightarrow& t \:<\:3\end{array}\quad\Rightarrow\quad t \in (1,3)$

Back-substitute: . $\begin{array}{ccccccc}2^x-2\:=\:1 & \Rightarrow& 2^x \:=\:3 &\Rightarrow& x \:=\:\log_23 \\ 2^x-2 \:=\:3 &\Rightarrow& 2^x \:=\:5 &\Rightarrow& x \:=\:\log_25 \end{array}$

Therefore: . $\log_23 \:<\:x\:<\:\log_25$

• July 12th 2009, 02:01 PM
Logic
Yes, yes, it is greater than.

Your solution seems very nice, too, however this is not the full answer.
It is said that every number from the interval minus infinity to zero also works, and, indeed, for -1, for example, I get 24 > 21 and -1 is not from $

\log_23 \:<\:x\:<\:\log_25
$
right.. ?