# Can someone help me solve this?

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#1
So I got a maths question that is:

Solve this equation
x ^-2/3 = 3x^-1

But I'm a bit stuck so can anyone help me answer this?
0
4 weeks ago
#2
(Original post by mellow2006)
So I got a maths question that is:

Solve this equation
x ^-2/3 = 3x^-1

But I'm a bit stuck so can anyone help me answer this?
you could cube both sides, leaving x^-2 = 27x^-3 . Take the reciprocals and then factorise. hope this helps
Last edited by sillysillyman; 4 weeks ago
1
#3
(Original post by sillysillyman)
you could cube both sides, leaving x^-2 = 3x^-3 . Take the reciprocals and then factorise. hope this helps
Ok that helps but how would I factorise the reciprocals?
0
4 weeks ago
#4
(Original post by mellow2006)
Ok that helps but how would I factorise the reciprocals?
If youre cubing both sides, you need to cube the whole of the right hand side.
Factorizing reciprocals is the same as normal, pull out the common term. You could write them as fractions to make it clearer.
1
4 weeks ago
#5
(Original post by mellow2006)
Ok that helps but how would I factorise the reciprocals?
sorry, I wasn't clear at all. raise both sides by ^-3 (similar to cubing and then ^-1(taking reciprocal)). this would leave you with
(x^-2/3)^-3=(3x^-1)^-3
x^2=(3^-3)x^3 (we multiply the powers)
x^2=(x^3)/27 then take the terms to one side and equal to zero, then factorise to get the solutions.
0
4 weeks ago
#6
(Original post by sillysillyman)
sorry, I wasn't clear at all. raise both sides by ^-3 (similar to cubing and then ^-1(taking reciprocal)). this would leave you with
(x^-2/3)^-3=(3x^-1)^-3
x^2=(3^-3)x^3 (we multiply the powers)
x^2=(x^3)/27 then take the terms to one side and equal to zero, then factorise to get the solutions.
Just to be clear, you realize there is only one solution and while factorising is generally the right thing to do, its possibly misleading here?
Last edited by mqb2766; 4 weeks ago
0
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