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How do you solve $$3\log(x-15)=\left(\frac{1}{4}\right)^x?$$ The solution is approximately $16$. How would you solve a logarithmic equation with an solution approximately equal to a number without using a graphing calculator?

dfeuer
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linksku
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    Once upon a time, men didn't have cellphones but were smart. They knew a continuous function that changes sign should intersect the x-axis somewhere in between. They would notice that if you make $x=16$ the left hand side is zero while the right hand side stays above zero. They would then put $x=17$ and notice the left hand side is $3\log(2)=\log(8)>1$, while $1/4^{17}<1$. – OR. Dec 09 '13 at 01:11
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    Men did not have cellphones? You are joking, right? – Igor Rivin Dec 09 '13 at 01:28
  • No, it is true. (http://en.wikipedia.org/wiki/Mobile_phone#History) – OR. Dec 09 '13 at 01:35
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    Only women had cell phones then. – dfeuer Dec 09 '13 at 01:39
  • But really, it seems most unlikely to have a nice answer at all. Comparing a logarithm to an exponential will not generally turn out well, and the logarithm of a sum does not make it any prettier. – dfeuer Dec 09 '13 at 01:40
  • This type of logarithmic and/or exponential equations can be expressed analytically in terms of the Lambert W function. If you want, I could post a detailed answer later, showing how this is done. – Lucian Dec 09 '13 at 05:33
  • @dfeuer Men not having cellphones doesn't imply women having cellphones. $(M\nsubseteq C)\nRightarrow(M^{c}\subset C)$. – OR. Dec 10 '13 at 18:19
  • @Lucian I would like to see the solution using Lambert function. To me, this case seems quite non-trivial for Lambert function. – pisoir Feb 07 '14 at 01:01
  • @pisoir: You're right. This one can't even be expressed in terms of Lambert's function, since it's the equivalent of $(\ln t)e^t=a$, as opposed to $te^t=a$. – Lucian Feb 07 '14 at 01:17

2 Answers2

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You don't say what base of logs you are using. The approach will be the same in any case-I will assume natural logs. We must have $x \gt 15$ or the logarithm is not defined. In that case, the right side will be very small and positive. We need $x \gt 16$ to make the left side positive. Define $a=x-16$, where we expect $a$ to be very small, so we will use the first term of the Taylor series. $$3 \log (x-15)=\left(\frac 14\right)^{\!x}\\ 3 \log (1+a)=\left(\frac 14\right)^{\!16+a}\\ 3\cdot 4^{16}a=4^{-a}$$ This shows $a \approx \frac 1{3\cdot 4^{16}}$ with both sides very close to $1$, so $x \approx 16+\frac 1{3\cdot 4^{16}}$.

dfeuer
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Ross Millikan
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The point is that for $x>10$ the right hand side is smaller than $1/1000000,$ so you are essentially solving for $LHS = 0.$ There is no general method if you are not so lucky.

Igor Rivin
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