How do you solve #2x^(3/4)=54#?

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Answer 1 · 2016-07-31T08:57:30

#x = 81#

Divide both sides by #2# to get:

#x^(3/4) = 27 = 3^3#

If #x# is Real and positive then so is #x^(3/4)# and we can raise both sides to the power #4/3# to find:

#x = x^1 = x^(3/4*4/3) = (x^(3/4))^(4/3) = (3^3)^(4/3) = 3^(3*4/3) = 3^4 = 81#

#color(white)()#
Other solutions?

Are there any negative or Complex solutions?

Suppose #x = r(cos theta + i sin theta)# where #r > 0# and #theta in (-pi, pi]#

Then:

#x^(3/4) = r^(3/4)(cos ((3theta)/4) + i sin ((3 theta)/4))#

For the imaginary part to be #0#, we must have #sin ((3theta)/4) = 0#

So #theta = (4pi)/3 k# for some integer #k#

This can only lie in the range #(-pi, pi]# if #k = 0#, so #theta = 0#.

So the only solution is the one on the positive part of the Real axis, i.e. #81#.