How do you solve 'unknown power' in an equation

[edmondng]

Title is a bit off but i don't know what it is called.

Basically,
If you have something like

3.35*10^10 = (400)^(4-x) - (300)^(4-x) and you want to solve for x.

Putting inside TI89, i get the answer x = -0.106072

But what if you do not have any calculator. I'm just wondering because i have a lot of numbers to solve and want to do it in software. I would have to import a lot of data, and each data is independent from each other. Doing manually would work but will take weeks.

edit: ok i took the natural log both sides.
would it come out like this: ln(3.35*10^10) = (4-x)ln(400) - (4-x)ln(300). actually what happens when you have a minus? a ln of division results in minus but how about minus before ln

Thanks

 

[sutupidmath]

lnx-lny=ln(\frac{x}{y}) in your problem try to factor out a 4-x so you are left with (4-x)ln(\frac{4}{3})

 

[edmondng]

actually my ln is wrong.
(400)^(4-x) - (300)^(4-x) != (4-x)ln(400) - (4-x)ln(300)

2^2 + 3^2 != e^(2ln2 + 2ln3) but (2^2)*(3^2) = e^(2ln2 + 2ln3)

 

[FrogPad]

The 89 loves algebra. Try to work it out by hand, but when you get stuck (or forget a rule) you could try something like:

solve(a=b^y-c^y,y)

where y=4-x, etc...

 

[sutupidmath]

what do you mean your ln is wrong? ln(3.35*10^10) = (4-x)ln(400) - (4-x)ln(300). there is nothing wrong with this

 

[edmondng]

what do you mean your ln is wrong? ln(3.35*10^10) = (4-x)ln(400) - (4-x)ln(300). there is nothing wrong with this

my original statement was
3.35*10^10 = (400)^(4-x) - (300)^(4-x)

i took the natural log of both sides. I thought it will be
ln(3.35*10^10) = (4-x)ln(400) - (4-x)ln(300)

but instead it should be
ln(3.35*10^10) = ln[400^(4-x) - 300^(4-x)] hence my original statement is already wrong.

I do not want to use the calculator because i have a lot more of these data all independent of each other which i would like to solve using software. I rather not use the calculator to keep solving for x because it would take weeks, and if data changes round and round it goes again. My data is in excel. i can import into java, or maybe excel can solve it..?

Thanks

 

[sutupidmath]

my original statement was
3.35*10^10 = (400)^(4-x) - (300)^(4-x)

i took the natural log of both sides. I thought it will be
ln(3.35*10^10) = (4-x)ln(400) - (4-x)ln(300)

but instead it should be
ln(3.35*10^10) = ln[400^(4-x) - 300^(4-x)] hence my original statement is already wrong.

I do not want to use the calculator because i have a lot more of these data all independent of each other which i would like to solve using software. I rather not use the calculator to keep solving for x because it would take weeks, and if data changes round and round it goes again. My data is in excel. i can import into java, or maybe excel can solve it..?

Thanks

Yeah, i see you are right, i did not look at the original problem at all! My bad!

 

[edmondng]

no problem. I'm just wondering what would be my best approach to solve for x or how would one solve it manually

 

[arildno]

You won't be able to find a nice exact solution to this, but some numerical approach, like Newton-Raphson would work.

 

[edmondng]

thanks
Newton's method work. i tried initially by inserting small increments and comparing how close it is to zero but takes too long because of the range. with Newton takes like seconds

its a little off from the calculator or excel calculations but could be due to rounding errors in the function

 

[edmondng]

here's an issue i am facing. when i use the calculator, it will solve and give me 2 or 3 answer. Normally 3. I know my answer has to be positive, so i can remove 1 of them. Now left with 2. I also know if i substitute this answer to find the other unknown, i can know which answer i need to choose based on my other unknown.

With Newton method, it works only so well as long you pick the right value for it to converge. Also it returns only 1 value, and i check my data few of them are wrong.

Any other ideas? I like to include conditional commands in my code to accommodate additional solutions then do some calculation/comparison to find the correct value but am not sure since with Newton it returns only 1. Of course i could run iterations with small increments but will take a long time.

Thanks

 

[helpneeded]

I have a mind teaser there: NO CALCULATORS..gotta solve this equation:

(1/5)^m. (1/4)^18 = [1/(2((10)^35))]
m=?

 

[arildno]

Well, since (\frac{1}{4})^{18}=(\frac{1}{2^{2}})^{18}==(\frac{1}{2})^{36},\frac{1}{2(10)^{35}}=(\frac{1}{2})^{36}(\frac{1}{5})^{35}, my brain was not much teased.