First of all, let's simplify a bit:
As e^x and ln(x) are inverse functions, e^ln(x) = x holds as well as ln(e^x) = x. This means that you can simplify your third logaritmic term:
log_8(1-x) + (10 log_32(x)) / 3 - log_2((1/x)/3) = 4/3
<=> log_8(1-x) + (10 log_32(x)) / 3 - log_2(1/(3x)) = 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(x) = log_b(x) / log_b(a)
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(1-x) + (10 log_32(x)) / 3 - log_2(1/(3x)) = 4/3
<=> (log_2(1-x))/(log_2(8)) + (10 log_2(x)) / (3 log_2(32)) - log_2(1/(3x)) = 4/3
Now, we can calculate log_2(8) = 3 and log_2(32) = 5
(in case it's not clear let me break it down just to be sure: log_2(8) = x <=> 2^(log_2(8)) = 2^x <=> 8 = 2^x <=> 2^3 = 2^x )
This leads us to the following, simpler, logarithmic equation:
(log_2(1-x))/3 + (10 log_2(x)) / (3*5) - log_2(1/(3x)) = 4/3
<=> 1/3 log_2(1-x) + 2/3 log_2(x) - log_2(1/(3x)) = 4/3
... multiply both sides with 3...
<=> log_2(1-x) + 2 log_2(x) - 3 log_2(1/(3x)) = 4
Now we are ready to use the logarithm rules:
log_a(x * y ) = log_a(x) + log_a(y) and log_a(x^y) = y * log_a(x)
The goal is to have just one log term at the left side. Let's do it. :)
log_2(1-x) + 2 log_2(x) - 3 log_2(1/(3x)) = 4
<=> log_2(1-x) + log_2(x^2) + log_2((1/(3x))^(-3)) = 4
<=> log_2(1-x) + log_2(x^2) + log_2(27 x^3) = 4
<=> log_2((1-x) * x ^2 * 27 x^3) = 4
<=> log_2(27 x^5 - 27 x^6) = 4
At this point, we can get rid of the log_2(a) by applying the inverse function 2^a to both sides of the equation.
log_2(27 x^5 - 27 x^6) = 4
<=> 2^(log_2(27 x^5 - 27 x^6)) = 2^4
<=> 27 x^5 - 27 x^6 = 2^4
<=> 27 x^5 - 27 x^6 = 16
<=> -x^6 + x^5 = 16/27
<=> -x^6 + x^5 - 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(x) = - x^6 + x^5 - 16/27 tells me that this equation has no solutions in RR.
graph{- x^6 + x^5 - 16/27 [-9.63, 10.37, -4.88, 5.12]}
I hope that this helped a little bit!