Hi All,

if I have a function

y = (x-2)^5

how can I express it in terms of y, ie

x = (some magical thing with y)

Thanks for your help

Michael

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- August 27th 2005, 10:19 PMmichael5139simple exponential question
Hi All,

if I have a function

y = (x-2)^5

how can I express it in terms of y, ie

x = (some magical thing with y)

Thanks for your help

Michael - August 27th 2005, 11:00 PMticbol
In other words you want to solve for x in terms of y.

In general, if you are finding the value of a variable, do anything or everything to isolate that variable---to let it stand all alone by itself in one side of the equation or inequality.

y = (x-2)^5

Here you want to find x. This x is in the rightside of the equation but it is not alone there. So let us isolate it.

First we remove/clear/eliminate the power or exponent of 5. We have to make the "5" into "1" only. How? We divide the "5" by itself, by 5 also. Because 5/5 = 1. How to do that?

We raise the whole (x-2)^5 to its 1/5 power. (That is the same as getting the 5th root of the thing.). But then, we do that on the (x-2)^5, so we have to do it also on the "y". We have to do exactly the same thing to both sides of the equation so that the equality will remain. (You know that already, I am sure.)

Thus,

y^(1/5) = [(x-2)^5]^(1/5)

y^(1/5) = (x-2)^(5/5)

y^(1/5) = (x-2)^1

y^(1/5) = x -2

Here x is still with a -2 on the right side. We eliminate that -2 so that x will be isolated by itself. How? We add 2 to both sides,

y^(1/5) +2 = x ---------***

There, x is all alone in the right side. Hence, that can be the answer, or,

x = y^(1/5) +2 ------answer.

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It does not have to be this long when you already know how to do it someday.