Solving inequality with logarithm

**jut** · December 11th 2009, 12:38 PM

I'm trying to solve this inequality for N where c, p, a are constants. For the 3rd equation, I flipped the inequality because I multiplied both sides by -1. But I'm not sure if the inequality changes when taking the log of both sides?

$10\log \left(\frac{1}{1+\left(\frac{c}{p}\right)^{2N}}\ri ght)>-a$

$-10\log \left(1+\left(\frac{c}{p}\right)^{2N}\right)>-a$

$10\log \left(1+\left(\frac{c}{p}\right)^{2N}\right)<a$

$\log \left(1+\left(\frac{c}{p}\right)^{2N}\right)<\frac {a}{10}$

exponentiate both sides...

$\left(\frac{c}{p}\right)^{2N}<10^{a/10}-1$

take log of both sides...

$2N \log \left(\frac{c}{p}\right)<\log \left(10^{a/10}-1\right)$

$N<\frac{\log \left(10^{a/10}-1\right)}{2 \log \left(\frac{c}{p}\right)}$

**abender** · December 11th 2009, 02:59 PM

Suppose we have this:

Greater Number > Lesser Number

Then we also have this:

log(Greater Number) > log(Lesser Number)

Taking the log of each side will not switch the inequality sign. Log(greater number) will still be greater than log(lesser number). What logging each side will do is cut down the value difference between the lesser side and greater side. Example: 1000>>0 but 6.9078 = log(1000) > log(10) =2.3026.

Note that >> means "much much greater than" as opposed to simply "greater than"

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