How do you solve 'unknown power' in an equation

  • Thread starter edmondng
  • Start date
  • Tags
    Power
In summary, the conversation discusses different methods of solving a mathematical problem involving finding the value of x in an equation. The methods discussed include using a calculator, using software, and solving it manually. The conversation also touches on the idea of finding multiple solutions and choosing the correct one, as well as the use of conditional commands to accommodate for these solutions.
  • #1
edmondng
159
0
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
 
Last edited:
Physics news on Phys.org
  • #2
[tex]lnx-lny=ln(\frac{x}{y})[/tex] in your problem try to factor out a 4-x so you are left with [tex](4-x)ln(\frac{4}{3})[/tex]
 
  • #3
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)
 
  • #4
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...
 
  • #5
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
 
  • #6
sutupidmath said:
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
 
  • #7
edmondng said:
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!
 
  • #8
no problem. I'm just wondering what would be my best approach to solve for x or how would one solve it manually
 
  • #9
You won't be able to find a nice exact solution to this, but some numerical approach, like Newton-Raphson would work.
 
  • #10
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
 
  • #11
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
 
  • #12
I have a mind teaser there: NO CALCULATORS..gotta solve this equation:

(1/5)^m. (1/4)^18 = [1/(2((10)^35))]
m=?
 
  • #13
Well, since [tex](\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}[/tex], my brain was not much teased.
 

Related to How do you solve 'unknown power' in an equation

1. How do you solve for an unknown power in an equation?

To solve for an unknown power in an equation, you can use logarithms or exponents. Logarithms can be used to solve for the unknown power when it is in the exponent position. Exponents can be used to solve for the unknown power when it is in the base position.

2. Can you provide an example of solving for an unknown power using logarithms?

Yes, for example, if we have the equation 2^x=16, we can rewrite it as log2(16)=x. Using a calculator, we can solve for x by taking the logarithm of both sides.

3. How do you solve for an unknown power when it is in the denominator?

To solve for an unknown power when it is in the denominator, you can use the reciprocal rule. This means that you can rewrite the equation as the power in the numerator and the base in the denominator. For example, if we have the equation 2^(1/x)=4, we can rewrite it as 1/x=log2(4). Then, solve for x by taking the reciprocal of both sides.

4. Is there a general formula for solving for an unknown power?

Yes, the general formula for solving for an unknown power is log(base)^(result)=power. This formula can be applied to different types of equations, as long as the power is the only unknown variable.

5. How do you check if your solution for an unknown power is correct?

To check if your solution for an unknown power is correct, you can substitute the value you found back into the original equation and see if it satisfies the equation. If it does, then your solution is correct. You can also use a calculator to verify your solution.

Similar threads

Replies
1
Views
326
Replies
8
Views
1K
  • Calculus
Replies
4
Views
2K
Replies
3
Views
1K
Replies
12
Views
2K
Replies
3
Views
925
  • Calculus
Replies
1
Views
1K
Back
Top