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

1 Answer
Nov 11, 2015

Answer:

No solutions in #RR#.

Explanation:

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!