#8x=4^x#.. Find x?

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Answer 1 · 2017-06-20T08:35:50

There are two solutions: #x=2, x~=0.18#

Most of the time, questions of this nature are quite difficult. However, by observation, we can see that when #x=2#:

#8(2)=16=4^2#

Is it possible that there is more than one solution? Yes.

And so another way to approach these kinds of questions is to graph the right side and the left side separately:

graph{(y-8x)(y-4^x)=0[-2,5,-5,20]}

By the graph, we can see that there is a second solution that is approximately #x=0.18#

Answer 2 · 2017-06-20T09:40:27

#x = {0.15495346619034528, 2}#

Introducing the Lambert function #W#

we have

#X = Y e^Y hArr Y = W(X)#

so

#8x = 4^x rArr 4* 2x=2^(2x)#

Making now #xi=2x# we have

#4 xi = 2^xi = (e^(log 2))^xi = e^(xi log 2)# or

#1/4 = xi e^(-xi log 2)#

multiplying both sides by #-log 2# we have

#-log 2/4 = -(xi log 2)e^(-xilog 2)#

now making #X=-log 2/4# and #Y = -xi log 2# we have

#-xi log2=W(-log 2/4)# or

#xi = -1/log 2 W(-log 2/4) = 2x# then

#x = -1/(2 log 2)W(-log 2/4) = (0.15495346619034528, 2)#