problem finding the exponent x in equation 0.920301^x + 0.17414^x = 1

mohit.sangavikar

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Apr 11, 2018
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this is not homework.
I am trying to find a exponent that fits my curve, in the end my equation boils down to this.

0.920301^x+0.17414^x=1

and I have to find the x.
How do I solve this, I cant take Log since that doesnt simplify things.

alternately if you know how I can solve this in Matlab it will be helpful too.
Much appreciated!
 
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this is not homework.
I am trying to find a exponent that fits my curve, in the end my equation boils down to this.

0.920301x+0.17414x=1

and I have to find the x.
How do I solve this, I cant take Log since that doesnt simplify things.

alternately if you know how I can solve this in Matlab it will be helpful too.
Much appreciated!
Does your equation look like this:

\(\displaystyle \displaystyle{0.920301^x \ + \ 0.17414^x \ = \ 1}\)

If that is correct - then the proper way to write that equation in ASCII format is:

0.920301^x + 0.17414^x = 1

Please check/edit your OP accordingly.
 
I am trying to find a exponent that fits my curve, in the end my equation boils down to this.

0.920301^x+0.17414^x=1

and I have to find the x.
What method(s) have you been taught for dealing with this sort of equation? How have you applied that information?

Thank you! ;)
 
Since log(A+ B) is NOT log(A)+ log(B), nor is it any simple combination of log(A) and log(B) you are not going to be able to find a simple formula giving x.

I recommend some simple numeric method. For example, graphing \(\displaystyle y= 0.920301^x+0.17414^x\) (on https://www.desmos.com/calculator) I see that the graph crosses y= 1 for x just a little more that 1.

So I calculate \(\displaystyle y= 0.920301^1+0.17414^1= 0.920301+ 0.17414= 1.07715 \), a little larger than 1 and that \(\displaystyle y= 0.920301^{2}+0.17414^{2}= 0.877278670201\) which is less than 1. Since this is a continuous function, there must be a value of x between 1 and 2 such that the value is 1. We don't know where, exactly, that value is but the simplest thing to do is try half way between, 1.5. \(\displaystyle y= 0.920301^{1.5}+0.17414^{1.5}= 0.9555\). That is a little less than 1 so we now know that there is a solution between 1 and 1.5. Try half way, 1.25. \(\displaystyle 0.920301^{1.25}+ 0.17414^{1.25}= 0.99276198975315573696619546651512\), still less than 1 (but a lot closer!) so there is a solution between 1 and 1.25. Continue that way until you get a sufficiently close solution.
 
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