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Mathematics 15 Online
OpenStudy (rane):

solve ( w - 3) ^3 = 16

OpenStudy (rane):

@Nurali @Hero @mathstudent55

OpenStudy (anonymous):

with order of operations, you can first take out the exponent from the parenthesis area

OpenStudy (rane):

ik that can u pls do it so I can see how u did it my answer is wrong

OpenStudy (rane):

@atlas

OpenStudy (anonymous):

w - 3 = 16^(1/3) w = 16^(1/3) + 3

OpenStudy (atlas):

hkim is right

OpenStudy (anonymous):

let me clarify real quick with the last part w = [16^(1/3)] + 3

OpenStudy (atlas):

if two sides are equal you can perform the same operation on both sides without changing the equality - right?

OpenStudy (atlas):

so we take cube root on both sides to get: w-3 = 16^(1/3)

OpenStudy (rane):

yh I know but hits confusing to find 16^1/3 without calculator. can u tell me how to do it if u know

OpenStudy (atlas):

now you can find w = 3 + 16^(1/3) I added 3 on both sides

OpenStudy (anonymous):

it is an irrational sum... 16^(1/4) would be easy but 16^(1/3) isn't

OpenStudy (rane):

but im not suppose to use calculator as im instructed by my teacher

OpenStudy (atlas):

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

OpenStudy (atlas):

if you remember cube root of 2........you can find 16^(1/3)

OpenStudy (atlas):

or if you are free to use logarithmic tables

OpenStudy (solomonzelman):

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

OpenStudy (solomonzelman):

So \[\huge\color{red}{16^{\frac{1}{3}}=\sqrt[3]{16}=\sqrt[3]{(2^3) \times 2}=2\sqrt[3]{2}}\]

OpenStudy (solomonzelman):

w = 3+16^(1/3) would be exactly, \[\huge\color{red}{w=3+2\sqrt[3]{2}}\]

OpenStudy (rane):

thank you everyone :)

OpenStudy (solomonzelman):

Anytime, yw!

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