# Determine the a value 0 < a < 1 in 1/16 log_a256^("colog"_(a^2)256^(log_(a^4)256^{cdots^("colog"_(a^(2^64))256)}))="Im"[z] where z is the solution for 2^4033 z^2-2^2017 z+1 = 0 (A)1/4, (B)1/8, (C)1/16, (D)1/32, (E) 1/64 ?

Feb 3, 2018

The only answer I can give you is assuming (based on the lack of parentheses) that everything to the left each $\log$ is completely INSIDE of the logarithm.

I am not sure how to solve it otherwise, without using mathematics way beyond what is considered "Algebra".

The "solutions" are $a = 1.2228 \text{ " or " } 10.8925$

#### Explanation:

First, solving for $z$:

${2}^{4033} {z}^{2} - {2}^{2017} z + 1 = 0$

$z = \frac{{2}^{2017} \pm \sqrt{{2}^{2 \cdot 2017} - 4 \left(1\right) \left({2}^{4033}\right)}}{2 \left({2}^{4033}\right)}$

$z = \frac{{2}^{2017} \pm \sqrt{{2}^{4034} - {2}^{4035}}}{2} ^ 4034$

$z = \frac{{2}^{2017} \pm \sqrt{- {2}^{4034}}}{2} ^ 4034$

$z = \left({2}^{-} 2017 \pm {2}^{-} 2017 i\right)$

$\therefore I m \left[z\right] = \pm {2}^{-} 2017$

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${\text{colog}}_{a} \left(x\right) = - {\log}_{a} \left(x\right)$

${\log}_{{a}^{b}} \left(x\right) = \frac{1}{b} {\log}_{a} \left(x\right)$

Also, your pattern of logs and cologs doesn't make sense, since it should end up with the log_(a^(2^64) being a normal log instead of a colog. For the purposes of this question, I will assume it was a typo and supposed to be a normal logarithm.
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Now, onto simplifying this horrific logarithm power tower (the math formatting literally cannot handle an expression of this size, so excuse the poor notation):

$\frac{1}{16} {\log}_{a} {256}^{{\text{colog}}_{{a}^{2}} {\ldots}^{\log} _ {a}^{{2}^{64}} \left(256\right)}$

We can pull the exponent out of the logarithm:

$= \frac{1}{16} {\text{colog}}_{{a}^{2}} {256}^{{\ldots}^{\log} \left({a}^{{2}^{64}}\right) 256} \cdot {\log}_{a} \left(256\right)$

And again:

$= \frac{1}{16} {\log}_{{a}^{4}} \left({256}^{{\ldots}^{\log} \ldots}\right) \cdot {\text{colog}}_{{a}^{2}} \left(256\right) \cdot {\log}_{a} \left(256\right)$

At this point, you can probably see where this is going. Each time we pull the exponent out of the logarithm, there's another exponent in THAT logarithm that can be pulled out too. So we can basically unravel this entire chain of logarithms into a simple product:

$= \frac{1}{16} {\log}_{a} 256 \cdot {\text{colog"_(a^2)256 * log_(a^4)256* "colog}}_{{a}^{8}} 256. . . \cdot {\log}_{{a}^{{2}^{64}}} 256$

Now, let's turn all of the cologs into negative logs:

$= \frac{1}{16} {\log}_{a} 256 \cdot \left(- {\log}_{{a}^{2}} 256\right) \cdot {\log}_{{a}^{4}} 256 \cdot \left(- {\log}_{{a}^{8}} 256\right) \ldots \cdot {\log}_{{a}^{{2}^{64}}} 256$

The negative signs occur in every other term, and there are 65 terms. Therefore, there will be 32 negative signs. Since this is an even number, we can just cancel out all the negative signs (using the fact that ${\left(- 1\right)}^{32} = 1$)

$= \frac{1}{16} {\log}_{a} 256 \cdot {\log}_{{a}^{2}} 256 \cdot {\log}_{{a}^{4}} 256 \cdot {\log}_{{a}^{8}} 256. . . \cdot {\log}_{{a}^{{2}^{64}}} 256$

Now that we just have a product of a bunch of logs, we can use the second rule from above, to rewrite all of the logs into base $a$.

$= \frac{1}{16} {\log}_{a} 256 \cdot \left(\frac{1}{2} {\log}_{a} 256\right) \cdot \left(\frac{1}{4} {\log}_{a} 256\right) \cdot \left(\frac{1}{8} {\log}_{a} 256\right) \ldots \cdot \left(\frac{1}{2} ^ 64 {\log}_{a} 256\right)$

$= \frac{1}{16} \left(\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{8.} . . \cdot \frac{1}{2} ^ 64\right) {\left({\log}_{a} \left({2}^{8}\right)\right)}^{65}$

$= \frac{1}{16} \left(\frac{1}{2} ^ \left(1 + 2 + 3. . . + 64\right)\right) {\left(8 {\log}_{a} 2\right)}^{65}$

$= \frac{1}{16} \left(\frac{1}{2} ^ 2080\right) {8}^{65} {\left({\log}_{a} 2\right)}^{65}$

$= \frac{1}{2} ^ 2096 {\left({2}^{3}\right)}^{65} {\left({\log}_{a} 2\right)}^{65}$

$= \frac{1}{2} ^ 2096 \left({2}^{195}\right) {\left({\log}_{a} 2\right)}^{65}$

$= \frac{1}{2} ^ 1901 {\left({\log}_{a} 2\right)}^{65}$

Now, we can set this equal to the $I m \left[z\right]$ we found earlier:

$\frac{1}{2} ^ 1901 {\left({\log}_{a} 2\right)}^{65} = \pm \frac{1}{2} ^ 2017$

${\left({\log}_{a} 2\right)}^{65} = \pm \frac{1}{2} ^ 116$

${\log}_{a} 2 = {2}^{\pm \frac{116}{65}}$

2 = a^(2^(+-116/65)

${2}^{\frac{1}{{2}^{\pm \frac{116}{65}}}} = a$

This produces the solutions $a = 1.2228$ and $a = 10.8925$, neither of which are in the range provided.

Like I said before, I'm not sure how else to help with this problem. If these solutions were what you were looking for, that's great, but if not, I would contact someone to ask them why such a mess of a problem was created.