solve ( w - 3) ^3 = 16
@Nurali @Hero @mathstudent55
with order of operations, you can first take out the exponent from the parenthesis area
ik that can u pls do it so I can see how u did it my answer is wrong
@atlas
w - 3 = 16^(1/3) w = 16^(1/3) + 3
hkim is right
let me clarify real quick with the last part w = [16^(1/3)] + 3
if two sides are equal you can perform the same operation on both sides without changing the equality - right?
so we take cube root on both sides to get: w-3 = 16^(1/3)
yh I know but hits confusing to find 16^1/3 without calculator. can u tell me how to do it if u know
now you can find w = 3 + 16^(1/3) I added 3 on both sides
it is an irrational sum... 16^(1/4) would be easy but 16^(1/3) isn't
but im not suppose to use calculator as im instructed by my teacher
normally we factorize the number and group the similar factors into pairs of 3.......but here 16 = 2*2*2*2....................there are four times 2 ....can't be grouped into 3
if you remember cube root of 2........you can find 16^(1/3)
or if you are free to use logarithmic tables
you know that there is a formula for 16^(1/3) \[\huge\color{red}{a^{\frac{b}{c}}=\sqrt[c]{a^b}}\] in your case, a=16 b=1 c=3
So \[\huge\color{red}{16^{\frac{1}{3}}=\sqrt[3]{16}=\sqrt[3]{(2^3) \times 2}=2\sqrt[3]{2}}\]
w = 3+16^(1/3) would be exactly, \[\huge\color{red}{w=3+2\sqrt[3]{2}}\]
thank you everyone :)
Anytime, yw!
Join our real-time social learning platform and learn together with your friends!