# Help using a Casio fx-83ES to calculate a fraction as an exponent.

• Jul 31st 2012, 02:32 AM
whammybird
Help using a Casio fx-83ES to calculate a fraction as an exponent.
I am trying to teach myself algebra and I have downloaded a course book, but I arrived at the section for inputting algebra in to graphic calculators (I only own the Casio fx-83ES model) I have figured out some of the basics but I have arrived at :[e.g. 2x^2 + 3y^3 / (x+y)^1/2] could anyone please tell me how to (if at all possible) enter a fraction as the exponent (^1/2), I realise this is very basic but even the Casio user guide isn't very clear on this.
• Jul 31st 2012, 02:39 AM
Prove It
Re: Help using a Casio fx-83ES to calculate a fraction as an exponent.
Quote:

Originally Posted by whammybird
I am trying to teach myself algebra and I have downloaded a course book, but I arrived at the section for inputting algebra in to graphic calculators (I only own the Casio fx-83ES model) I have figured out some of the basics but I have arrived at :[e.g. 2x^2 + 3y^3 / (x+y)^1/2] could anyone please tell me how to (if at all possible) enter a fraction as the exponent (^1/2), I realise this is very basic but even the Casio user guide isn't very clear on this.

Press the power button then enter the fraction in brackets.
• Jul 31st 2012, 02:56 AM
whammybird
Re: Help using a Casio fx-83ES to calculate a fraction as an exponent.
Attachment 24390 sorry for the terrible photo but this is what I am inputting in to the device, you can see the bracketed fraction shown as an exponent in the denominator. It is telling me there is a Math error when I try to get it to calculate. Thanks for the advice already and any in advance.
• Jul 31st 2012, 04:21 AM
kraj8995
Re: Help using a Casio fx-83ES to calculate a fraction as an exponent.
Just download its manual from internet or consult with those who does it.
• Jul 31st 2012, 04:30 AM
emakarov
Re: Help using a Casio fx-83ES to calculate a fraction as an exponent.
I don't have the calculator, but note that $(x+y)^{1/2}$ is defined only if x + y ≥ 0.