Nerpilis
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this problem is an excerpt from an explanation of a time wieghted performance method. I feel that if I can follow this part the rest of it will make sense. now i know the answer is .1259 but I'm a little fuzzy on how exactly they got that, and their 'step by step' seems to miss some steps. it may be partially due to my rustiness with ln. It is as follows:
3000 = 2500(1 + R)^(31/31) + 175(1 + R)^(15/31)
3000 = 2500(1)ln(1+R) + 175(15/31)ln(1+R)
here I thought if you applied the natural log you would have to do it to both sides of the equation, and why were the constants left out of it? but the explanation did not do so I will continue as such, please correct me if I'm wrong.
3000 = [ln(1+R)][2500 + (175)(15/31)]
1.160686427 = ln(1+R)
this is about where i don't know the proper way to solve this. any explanations about using natural logs in these types of situations are much appreciated.
3000 = 2500(1 + R)^(31/31) + 175(1 + R)^(15/31)
3000 = 2500(1)ln(1+R) + 175(15/31)ln(1+R)
here I thought if you applied the natural log you would have to do it to both sides of the equation, and why were the constants left out of it? but the explanation did not do so I will continue as such, please correct me if I'm wrong.
3000 = [ln(1+R)][2500 + (175)(15/31)]
1.160686427 = ln(1+R)
this is about where i don't know the proper way to solve this. any explanations about using natural logs in these types of situations are much appreciated.