I am so lost.. Can somebody help me how to solve this:

Determine X when:

2X^4,6 = 512x^0,6

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- Jun 15th 2011, 01:49 AMoliverbjSolve Power Equation
I am so lost.. Can somebody help me how to solve this:

Determine X when:

2X^4,6 = 512x^0,6 - Jun 15th 2011, 01:52 AMAckbeetre: Solve Power Equation
First thing I would do is divide both sides by 2. What do you get?

- Jun 15th 2011, 02:06 AMoliverbjre: Solve Power Equation
- Jun 15th 2011, 02:10 AMAckbeetre: Solve Power Equation
Correct. Now what do you suppose you could do next? You have x raised to a power on both sides of the equation. You'd like to isolate x on one side of the equation. Is there an operation you could do to both sides of the equation that will move you in that direction?

[EDIT]: You should only have divided the coefficients by two, not the exponents. - Jun 15th 2011, 02:14 AMoliverbjre: Solve Power Equation
- Jun 15th 2011, 02:17 AMAckbeetre: Solve Power Equation
You'd like to get rid of the exponents somehow. Do you know a function that converts exponents to something else?

Quote:

But a little question to the title; you've named it "Exponential Equation" - the subject is a "Power equation"?

- Jun 15th 2011, 02:25 AMProve ItRe: Solve Power Equation
- Jun 15th 2011, 02:45 AMoliverbjRe: Solve Power Equation
Wow, thanks a lot! When it is displayed like that, it seems very simple! I get it now. Thanks again.

But another question, I have to determine the regulation of the Power Equation, when the following two points, is on the power equations graph:

(2,4) and (4,32)

How in earth is that done? Or how do I just get started with it? :)

Thanks a lot in advance, what a great forum! - Jun 15th 2011, 04:20 AMHallsofIvyRe: Solve Power Equation
A "power equation" is of the form . If y= 4 when x= 2, then . If y= 32 when x= 4, .

Solve those two equations for A and k. (Trying eliminating A first by**dividing**one equation by the other.) - Jun 15th 2011, 09:34 AMskeeterRe: Solve Power Equation
- Jun 15th 2011, 10:13 AMHallsofIvyRe: Solve Power Equation
Right. What Prove It should have said was

**If x is not 0**

then

so the three real roots are 0, 4 and -4.

We could also note that so taking the negative root, gives the two imaginary roots .