# What is x if log_8(1-x) + (10log_32(x))/3-log_2(e^ln(1/x)/3)=4/3?

Nov 11, 2015

No solutions in $\mathbb{R}$.

#### Explanation:

First of all, let's simplify a bit:

As ${e}^{x}$ and $\ln \left(x\right)$ are inverse functions, ${e}^{\ln} \left(x\right) = x$ holds as well as $\ln \left({e}^{x}\right) = x$. This means that you can simplify your third logaritmic term:

${\log}_{8} \left(1 - x\right) + \frac{10 {\log}_{32} \left(x\right)}{3} - {\log}_{2} \left(\frac{\frac{1}{x}}{3}\right) = \frac{4}{3}$

$\iff {\log}_{8} \left(1 - x\right) + \frac{10 {\log}_{32} \left(x\right)}{3} - {\log}_{2} \left(\frac{1}{3 x}\right) = \frac{4}{3}$

Your next goal is to bring all the $\log$ functions to the same base so that you have a chance to use logarithm rules on them and simplify.

You can change the logarithm base as follows:
${\log}_{a} \left(x\right) = {\log}_{b} \frac{x}{\log} _ b \left(a\right)$

Let's use this rule to change the base $8$ of ${\log}_{8}$ and the base $32$ of ${\log}_{32}$ to base $2$:

${\log}_{8} \left(1 - x\right) + \frac{10 {\log}_{32} \left(x\right)}{3} - {\log}_{2} \left(\frac{1}{3 x}\right) = \frac{4}{3}$
$\iff \frac{{\log}_{2} \left(1 - x\right)}{{\log}_{2} \left(8\right)} + \frac{10 {\log}_{2} \left(x\right)}{3 {\log}_{2} \left(32\right)} - {\log}_{2} \left(\frac{1}{3 x}\right) = \frac{4}{3}$

Now, we can calculate ${\log}_{2} \left(8\right) = 3$ and ${\log}_{2} \left(32\right) = 5$
(in case it's not clear let me break it down just to be sure: ${\log}_{2} \left(8\right) = x \iff {2}^{{\log}_{2} \left(8\right)} = {2}^{x} \iff 8 = {2}^{x} \iff {2}^{3} = {2}^{x}$)

This leads us to the following, simpler, logarithmic equation:

$\frac{{\log}_{2} \left(1 - x\right)}{3} + \frac{10 {\log}_{2} \left(x\right)}{3 \cdot 5} - {\log}_{2} \left(\frac{1}{3 x}\right) = \frac{4}{3}$

$\iff \frac{1}{3} {\log}_{2} \left(1 - x\right) + \frac{2}{3} {\log}_{2} \left(x\right) - {\log}_{2} \left(\frac{1}{3 x}\right) = \frac{4}{3}$

... multiply both sides with $3$...
$\iff {\log}_{2} \left(1 - x\right) + 2 {\log}_{2} \left(x\right) - 3 {\log}_{2} \left(\frac{1}{3 x}\right) = 4$

Now we are ready to use the logarithm rules:

${\log}_{a} \left(x \cdot y\right) = {\log}_{a} \left(x\right) + {\log}_{a} \left(y\right)$ and ${\log}_{a} \left({x}^{y}\right) = y \cdot {\log}_{a} \left(x\right)$

The goal is to have just one $\log$ term at the left side. Let's do it. :)

${\log}_{2} \left(1 - x\right) + 2 {\log}_{2} \left(x\right) - 3 {\log}_{2} \left(\frac{1}{3 x}\right) = 4$
$\iff {\log}_{2} \left(1 - x\right) + {\log}_{2} \left({x}^{2}\right) + {\log}_{2} \left({\left(\frac{1}{3 x}\right)}^{- 3}\right) = 4$

$\iff {\log}_{2} \left(1 - x\right) + {\log}_{2} \left({x}^{2}\right) + {\log}_{2} \left(27 {x}^{3}\right) = 4$
$\iff {\log}_{2} \left(\left(1 - x\right) \cdot {x}^{2} \cdot 27 {x}^{3}\right) = 4$
$\iff {\log}_{2} \left(27 {x}^{5} - 27 {x}^{6}\right) = 4$

At this point, we can get rid of the ${\log}_{2} \left(a\right)$ by applying the inverse function ${2}^{a}$ to both sides of the equation.

${\log}_{2} \left(27 {x}^{5} - 27 {x}^{6}\right) = 4$

$\iff {2}^{{\log}_{2} \left(27 {x}^{5} - 27 {x}^{6}\right)} = {2}^{4}$

$\iff 27 {x}^{5} - 27 {x}^{6} = {2}^{4}$
$\iff 27 {x}^{5} - 27 {x}^{6} = 16$
$\iff - {x}^{6} + {x}^{5} = \frac{16}{27}$
$\iff - {x}^{6} + {x}^{5} - \frac{16}{27} = 0$

Unfortunately, I have to admit that I'm stuck at this moment since I don't know how to solve this equation.

However, plotting $f \left(x\right) = - {x}^{6} + {x}^{5} - \frac{16}{27}$ tells me that this equation has no solutions in $\mathbb{R}$.
graph{- x^6 + x^5 - 16/27 [-9.63, 10.37, -4.88, 5.12]}

I hope that this helped a little bit!