Solve Power Equation

• Jun 15th 2011, 01:49 AM
oliverbj
Solve 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 AM
Ackbeet
re: Solve Power Equation
First thing I would do is divide both sides by 2. What do you get?
• Jun 15th 2011, 02:06 AM
oliverbj
re: Solve Power Equation
Quote:

Originally Posted by Ackbeet
First thing I would do is divide both sides by 2. What do you get?

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

I will now divide by 2 on both sides:

X^2,3 = 256x^0,3

- Correct?
• Jun 15th 2011, 02:10 AM
Ackbeet
re: 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 AM
oliverbj
re: Solve Power Equation
Quote:

Originally Posted by Ackbeet
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?

I am not quite sure. We just divided on both sides.

But a little question to the title; you've named it "Exponential Equation" - the subject is a "Power equation"?
• Jun 15th 2011, 02:17 AM
Ackbeet
re: Solve Power Equation
Quote:

Originally Posted by oliverbj
I am not quite sure. We just divided on both sides.

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"?
Perhaps you're right. I'll rename again.
• Jun 15th 2011, 02:25 AM
Prove It
Re: Solve Power Equation
Quote:

Originally Posted by oliverbj
I am so lost.. Can somebody help me how to solve this:

Determine X when:

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

\displaystyle \displaystyle \begin{align*}2x^{4.6} &= 512x^{0.6} \\ x^{4.6} &= 256x^{0.6} \\ \frac{x^{4.6}}{x^{0.6}} &= 256 \\ x^{4.6 - 0.6} &= 256 \\ x^4 &= 256 \\ x &= 4 \end{align*}
• Jun 15th 2011, 02:45 AM
oliverbj
Re: Solve Power Equation
Quote:

Originally Posted by Prove It
\displaystyle \displaystyle \begin{align*}2x^{4.6} &= 512x^{0.6} \\ x^{4.6} &= 256x^{0.6} \\ \frac{x^{4.6}}{x^{0.6}} &= 256 \\ x^{4.6 - 0.6} &= 256 \\ x^4 &= 256 \\ x &= 4 \end{align*}

Quote:

Originally Posted by Prove It
\displaystyle \displaystyle \begin{align*}2x^{4.6} &= 512x^{0.6} \\ x^{4.6} &= 256x^{0.6} \\ \frac{x^{4.6}}{x^{0.6}} &= 256 \\ x^{4.6 - 0.6} &= 256 \\ x^4 &= 256 \\ x &= 4 \end{align*}

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 AM
HallsofIvy
Re: Solve Power Equation
A "power equation" is of the form $\displaystyle y= Ax^k$. If y= 4 when x= 2, then $\displaystyle 4= A(2^k)$. If y= 32 when x= 4, $\displaystyle 32= A(4^k)$.

Solve those two equations for A and k. (Trying eliminating A first by dividing one equation by the other.)
• Jun 15th 2011, 09:34 AM
skeeter
Re: Solve Power Equation
Quote:

Originally Posted by oliverbj
I am so lost.. Can somebody help me how to solve this:

Determine X when:

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

$\displaystyle 2x^{4.6} - 512x^{0.6} = 0$

$\displaystyle 2x^{0.6}(x^4 - 256) = 0$

$\displaystyle 2x^{0.6}(x^2+16)(x^2-16) = 0$

$\displaystyle 2x^{0.6}(x^2+16)(x+4)(x-4) = 0$

three real solutions ...
• Jun 15th 2011, 10:13 AM
HallsofIvy
Re: Solve Power Equation
Right. What Prove It should have said was

$\displaystyle x^{4.6}= 256x^{0.6}$
If x is not 0
then $\displaystyle \frac{x^{4.6}}{x^{0.6}}= x^4= 256$
$\displaystyle x= \pm 4$

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

We could also note that $\displaystyle x^2= \pm 16$ so taking the negative root, $\displaystyle x^2= -16$ gives the two imaginary roots $\displaystyle x= \pm 4i$.