rearranging power and log equations

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August 27th 2011, 06:39 AM
mathsonthemoor

rearranging power and log equations

Hello,

I have two equations that I want to reaarange, one a power, the other a log. I can solve the orignal equations but am getting stuck trying to make x the subject. Any help would be appreciated.

eq. 1: y= 4.3316x0.4671

To solve this (for x=8) I did the following: 4.3316 * 80.4671
= 8^(1/10 000) PWR 4671
= 4.3316 *2.64
= 11.44
I think this is right but don't know how to rearrange to get x= ?

Eq 2. y=9.5818ln(x) - 9.5878
Can solve this ok but stuck on same point - getting x=

Thanks
August 27th 2011, 06:56 AM
Siron

Re: rearranging power and log equations

Eq 2:
$y=9,5818\ln(x)-9,5878 \Leftrightarrow y+9,5878=9,5818\ln(x) \Leftrightarrow \frac{y+9,5878}{9,5818}=\ln(x) \Leftrightarrow x=e^{\frac{y+9,5878}{9,5818}}$

Eq 1:
This one is not clear in my opinion.
August 27th 2011, 07:44 AM
HallsofIvy

Re: rearranging power and log equations

I assume the problem (1) is to solve $y= 4.3316x^{0.4671}$ for x. (If you don't use LaTeX, at least use "^" to indicate powers.)

Start with the obvious: divide both sides by 4.3316: $\frac{y}{4.3316}= x^{0.4671}$
and then take the .4671 root of both sides: $x= \left(\frac{y}{4.3316}\right)^{1/0.4671}$ .

No logarithm need.

If the x had been in the exponent, then you would need the logarithm:
$y= 4.3316(0.4671)^x$
$\frac{y}{4.3316}= (0.4671)^x$
$log\left(\frac{y}{4.3316}\right)= log(0.4671^x)= xlog(0.4671)$
$x= \frac{log\left(\frac{y}{4.43316}\right)}{log(0.467 1)}$
August 27th 2011, 11:30 AM
mathsonthemoor

Re: rearranging power and log equations

Thanks for both your answers - it's so obvious when someone shows you how!

I also apologise for the rather sloppy nature of the post re. layout etc. Will make sure I'm clear if I need to post again.

Thanks again.
September 1st 2011, 09:11 AM
mathsonthemoor

Re: rearranging power and log equations

[QUOTE=Siron;675593]Eq 2:
$y=9,5818\ln(x)-9,5878 \Leftrightarrow y+9,5878=9,5818\ln(x) \Leftrightarrow \frac{y+9,5878}{9,5818}=\ln(x) \Leftrightarrow x=e^{\frac{y+9,5878}{9,5818}}$

Sorry to be stupid but I can't seem to get a sensible answer drom the above equation that you kindly showed me how to rearrange. When I solve the original equation for y using a series of values for x, I can't then use those same values of y to get the corresponding values of x that I used - I end up with insanely large values for x.

What I would give to be good at maths(Doh)
September 1st 2011, 11:22 AM
Siron

Re: rearranging power and log equations

I don't get your point, do you understand the steps I did or not? Can you give an example where you're stuck?

And it's not stupid, it's important to ask things what you don't understand.
September 2nd 2011, 11:26 AM
mathsonthemoor

Re: rearranging power and log equations

Ok. Here is my problem. Let's say in the original equation x =49. If I am correct this would give a value for y of 27.7. So now I want to use the rearranged equation, using this value of y (27.7), to get back to the original value of x I used (49) - to check I understand how to do it. This is where I am stuck. Looking at the equation, e is the base, which is ln or 2.718281828. Am I right here? y+9.5878/9.5818 is the power? This produces absurdly large anwers. I'm sure there is something pretty fundamental that I am not understanding?!
September 2nd 2011, 11:46 AM
Siron

Re: rearranging power and log equations

It's the same, indeed when $x=49$ then $y\approx 27,7028$ . When we substitutie this value for y in the equation:
$x=e^{\frac{y+9,5878}{9,5818}}$ then we get:
$x=e^{\frac{27,7028+9,5878}{9,5818}}$
$\Leftrightarrow x=48,998\approx 49$

So the rearranged equation is fully equivalent with the original one.
Does this answer your question?
September 3rd 2011, 09:32 AM
mathsonthemoor

Re: rearranging power and log equations

Yes thanks. Just realised what I've been doing wrong -trying to do the sum in the wrong order, which explains why I was getting it wrong and now it is correct. Thanks again.
September 3rd 2011, 12:35 PM
Siron

Re: rearranging power and log equations

You're welcome! :)