How do you solve #x^ { 9} = 135#?

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Answer 1 · 2018-01-07T05:52:50

#x=1.72466...#

#x^9=135#

#x=root(9)135#

Using a calculator, #root(9)135=1.72466...#

#:.x=1.72466...#

Answer 2 · 2018-01-07T10:34:51

#x = 135^(1/9)* e^( (2kpi i )/9 ) #

#k = { 0,1,2,3,4,5,6,7,8} #

Where #e^(itheta ) = costheta + i sintheta #

Alternate using complex numbers...
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# i = sqrt(-1) #

Where #e^(itheta ) = costheta + i sintheta #

But # k = {9, 10 , ... } # are repeated roots, so we dont include these

Answer 3 · 2018-01-07T10:48:34

A different approach. Method useful in other contexts. Which is why I am demonstrating it.

#x~~1.7246# to 4 decimal places.

Take logs of both sides (does nor matter which type of log)

#ln(x^9)=ln(135)#

#9ln(x)=ln(135)#

#ln(x)=ln(135)/9#

#x=ln^(-1)[ ln(135)/9]# in the very old days they called this antilog

#x~~1.7246# to 4 decimal places.

#color(brown)("You should always state the degree of rounding")#