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I have the following expression $$\frac{1}{1+\rho}(1+n)^{(1-\sigma)}*(1+\gamma_{A})^{1-\sigma}<1$$

and have to use logarithms to get the following

$$(1-\sigma)(n+\gamma_{A})<\rho$$

Could someone please offer me some help on how to do it? That's a homework question in Macroeconomics - we got the solutions and were told to use the logarithms to get them. However, I'm struggling with the complete procedure, so any help is more than welcome.

Thanks!

wakum
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  • what is it exactly you don't understand? – Alex Jan 29 '14 at 15:45
  • I am also having difficulties in getting your answer, that contains no logarithm whatsoever. But a logarithm was used somewhere to get that exponent down... – imranfat Jan 29 '14 at 16:01
  • Were you also given some sort of approximation like $\log(1+x)\approx x$? Because that appears to have happened several times here. – tabstop Jan 29 '14 at 16:07
  • @Alex, I don't understand how to simplify the first expression to get the second one. – wakum Jan 29 '14 at 17:11
  • @imranfat, yes we were instructed to use the algorithm to simplify the expression. Exactly, I was thinking of using the log to get rid of the exponent, I'm just not sure how to get there. – wakum Jan 29 '14 at 17:13
  • @tabstop, no, no approximation were specified ... – wakum Jan 29 '14 at 17:13
  • @wakum I sincerely believe that the simplication is then incorrect. There has to be a log term somewhere in order to get that exponent down. Anxiously waiting for an answer as well. – imranfat Jan 29 '14 at 17:16

1 Answers1

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Taking the logarithms brings the exponent down and turns all the multiplies into adds, so we have $$-\log(1+\rho) + (1-\sigma)\log(1+n) + (1-\sigma)\log(1+\gamma_A) < 0.$$ We can move the first term across and factor to get $$(1-\sigma)(\log(1+n)+\log(1+\gamma_A)) < \log(1+\rho).$$

If $\rho < 1$, then $\log(1+\rho) < \rho$ from the series expansion, so that will get us $\rho$ on the right. Unfortunately the same trick doesn't quite work for the left side, since I don't want to replace the left side with something bigger. Perhaps you could do something with changing the sum of logs into $\log(1+n+\gamma_A+n\gamma_A)$ and showing that that is bigger than $n+\gamma_A$ because of the extra bit at the end (again assuming $n$ and $\gamma_A$ are relatively small).

tabstop
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  • After checking numerically, it doesn't look like $\log(1+n+\gamma_A+n\gamma_A)>n+\gamma_A$ actually is true anywhere, so hmm. – tabstop Jan 29 '14 at 17:50
  • So I've been crunching some numbers in the background, and it would appear that we have some counterexamples; one picked more-or-less at random is $\rho=0.004$, $n=0.033$, $\sigma=0.894$, and $\gamma_A=0.005$. If my calculating is correct, the LHS of the original comes out to 0.99997818 (which is less than 1), and the LHS of the second version comes out to 0.004028 (which is bigger than $\rho$). So far the program has found 1478 counterexamples, and it's still chugging along. – tabstop Jan 30 '14 at 01:54
  • I suspect, given that for most small numbers $\log(1+x)\approx x$ isn't too bad, there may be some range where it works, but I don't know what bounds (if any) there are on these variables. – tabstop Jan 30 '14 at 01:56
  • Thanks, I'll mark this one as the answer for now. – wakum Feb 01 '14 at 22:37