$\displaystyle \log_{b-1}\frac{n}{b}=p$, where n and p are constants. Solve for b. Hmmm... Presumably I'd have to set both sides as the exponent of base (b-1) to eliminate the log. But then I have a cubic, right?
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Originally Posted by scorpion007 $\displaystyle \log_{b-1}\frac{n}{b}=p$, where n and p are constants. Solve for b. Hmmm... Presumably I'd have to set both sides as the exponent of base (b-1) to eliminate the log. But then I have a cubic, right? $\displaystyle (b - 1)^p = \frac{n}{b} \Rightarrow b(b - 1)^p - n = 0$. What happens next will depend on the values of p and n.
Last edited by mr fantastic; May 28th 2009 at 04:19 AM. Reason: Fixed a typo
Originally Posted by mr fantastic $\displaystyle (b - 1)^p = \frac{n}{p} \Rightarrow b(b - 1)^p - n = 0$. What happens next will depend on the values of p and n. $\displaystyle (b-1)^{p} = \frac{n}{b} $.
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