Solve for #x#? if #4=(1+x)^24#

3 Answers
Jun 16, 2018

#-1+2^(1/12)#

Explanation:

#4=(1+x)^24#

#root(24)4=1+x#

#4^(1/24)=1+x#

#2^(2/24)=1+x#

#2^(1/12)=1+x#

#-1+2^(1/12)=x#

Jun 17, 2018

Extend to complex numbers:

If anyone has studies complex numbers

Explanation:

#4 = (1+x)^24 #

# 4 = (1+x)^24 e^(2kpi i ) #

as #e^(2kpi i ) =1, AA k in ZZ #

#4^(1/24) = (1+x)e^(1/12 k pi i ) #

#=> 2^(1/12) = e^(1/12 k pi i) + xe^(1/12 k pi i ) #

#=> 2^(1/12) - e^(1/12 k pi i ) = xe^(1/12 k pi i) #

#=> x = (2^(1/12) - e^(1/12 k pi i ) ) / e^(1/12 k pi i ) #

#=> k = {0,1,2,3, ... , 22 , 23 } #

Jun 30, 2018

#x=2^(1/12)-1#

Explanation:

We can take the #24#th root of both sides to get

#4^(1/24)=1+x#

Subtracting #1# from both sides gives us

#x=4^(1/24)-1#

We can now rewrite #4# as #2^2#. This gives us

#x=2^(2*1/24)-1#

which can be simplified as

#x=2^(1/12)-1#

Hope this helps!