I am trying to solve the equation x^(2/3) = 1/16

Expert Answers
rakesh05 eNotes educator| Certified Educator

Here the equation is `x^(2/3)=1/16` .

Taking cube of both sides we get


or,        `(x)^((2/3).3)=(1/16)^3`

or,          `(x)^2=((1/4)^2)^3`

or,           `(x)^2=(1/4)^6`

`` or,            `(x)^2=((1/4)^3)^2`

or,          `x^2-((1/4)^3)^2=0`

or,        `(x-(1/4)^3)(x+(1/4)^3)=0`

So,      `(x-(1/4)^3)=0`  and `(x+(1/4)^3)=0`

so,         `x=(1/4)^3`   or     `x=-(1/4)^3`

So,     `x=1/64`       0r       `x=-1/64` .

violy eNotes educator| Certified Educator

Try to get rid of the exponent 2/3 on left side. To do that, we raised both sides by the reciprocal of 2/3 which is 3/2. 

`(x^(2/3))^(3/2) = (1/16)^(3/2)`

`x = (1/16)^(3/2)`

Take note that 3/2 = 1/2 * 3.

`x = ((1/16)^(1/2))^(3) = (1/4)^3`

The exponent indicate how many times we multiply the base by itself. 

`x = (1/4)^3 = (1/4)*(1/4)*(1/4) = 1/64` 

Hence, x = 1/64.

pramodpandey | Student






Factorise the equation by formula `a^2-b^2=(a-b)(a+b)` ,so we have











`` Thus




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