exponentials: solve 975^x / (975^x + 599^x) - 110/154 = 0

absolutjosh

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975^x / (975^x + 599^x) - 110/154 = 0, solve for x
 

stapel

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What methods have they covered recently in class? What have you tried?

Please be complete. Thank you! :D

Eliz.
 

absolutjosh

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this is actually the second part of a bigger problem. i can simplify the problem to the point of solving for the exponent. i have no idea how to do that.
 

stapel

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absolutjosh said:
this is actually the second part of a bigger problem....
Ah. So, since your class hasn't covered any techniques for solving this sort of equation, we probably need to see the rest of the exercise, from the beginning, to see if perhaps an error was made along the way. :idea:

. . . . .As the "Read Before Posting" thread says:

Post the complete text of the exercise. This would include the full statement of the exercise and its instructions, so the tutors will know what you are working on.... Even if you're asking only about the very end of the solution process, still include all of your intermediate steps. Errors may have occurred earlier than you'd realized....
Thank you! :D

Eliz.
 

absolutjosh

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The whole question is as follows:

Based on the number of runs scored (S) and runs allowed (A), the Pythagorean winning percentage estimates what a baseball team's winning percentage should be. This formula is (S^x)/(S^x + A^x). It has been determined that x=1.83 allows for the most accurate results. The 1927 New York Yankees are reguarded as one of the best teams in baseball history. Their record was 110 wins and 44 losses. They scored 975 runs while allowing only 599.

a. Find their Pythagorean win-loss record. My findings: 109 wins, 45 losses

b. Estimate the value of x to the nearest hundredth that best predicts the 1927 Yankees' actual win-loss record. Hint: It is not 1.83.

I've figured out to find this I have to plug in what i know which gives me (975^x)/(975^x+599^x) and then subtract their record, which is 110/154, from what i plugged in and that should give me 0. I just can't figure out how to solve for the x to find out what the value would be.
 

stapel

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Are you supposed to be using your graphing calculator for this? Or are you working with logarithms? Or are you doing numerical approximation of zeroes? Or something else?

I'll be glad to try to help you solve this (and I agree with your set-up), but I need some information regarding the method you are expected to use. For instance, what was the topic for the section in your textbook in which this homework question came up?

Thank you! :D

Eliz.

P.S. Their hint is correct. While close to 1.83, the value here is a smidge higher.
 

absolutjosh

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I'm in precal and we just finished an algebra review unit. The method of finding the solution doesn't matter. And I'm using a graphic calculator.
 

Denis

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You can't solve directly; you need to use "ITERATION"; google it
 

stapel

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absolutjosh said:
I'm in precal and we just finished an algebra review unit. The method of finding the solution doesn't matter. And I'm using a graphic calculator.
Have you covered logs yet? (I used logs to find the answer.)

If you're using your graphing calculator, then graph your equation and, after ZOOMing in a bit, have the calculator find an approximate x-intercept value.

If you're wanting more decimal places, do the approximation thing (narrowing down the choice by finding negative and positive answers that bracket the "zero" solution), which would have been covered back in algebra.

Eliz.
 

Subhotosh Khan

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Re: exponentials: solve 975^x / (975^x + 599^x) - 110/154 =

absolutjosh said:
975^x / (975^x + 599^x) - 110/154 = 0, solve for x
There is a closed form solution (amenable to pre-calc algebra)

975^x / (975^x + 599^x) - 110/154 = 0

975^x / (975^x + 599^x) = 110/154

(975^x + 599^x)/975^x = 1 + 44/110

(599/975)^x = 0.4

x * log(599/975) = log(0.4)

x * (-0.211577793) = -0.397940009

x = 1.880821245
 

Denis

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Denis said:
You can't solve directly; you need to use "ITERATION"; google it
Yikes...that's wrong, Denis! No need for iteration: thanks, Mr K :idea:

If a^x / (a^x + b^x) = u / v
then x = LOG[(v - u) / u] / LOG(b / a)

LOG[(154 - 110) / 110] / LOG(599 / 975) = 1.880821245....

Can be slightly changed:
If (a^x - b^x) / (a^x + b^x) = u / v
then x = LOG[(v - u) / (v + u] / LOG(b / a)
 
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