1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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 $1/4^{17}<1$. $\endgroup$ – OR. Dec 9 '13 at 1:11
  • 2
    $\begingroup$ Men did not have cellphones? You are joking, right? $\endgroup$ – Igor Rivin Dec 9 '13 at 1:28
  • $\begingroup$ No, it is true. (en.wikipedia.org/wiki/Mobile_phone#History) $\endgroup$ – OR. Dec 9 '13 at 1:35
  • 2
    $\begingroup$ Only women had cell phones then. $\endgroup$ – dfeuer Dec 9 '13 at 1:39
  • $\begingroup$ 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. $\endgroup$ – dfeuer Dec 9 '13 at 1:40
1
$\begingroup$

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 $x \approx 16+\frac 1{3\cdot 4^{16}}$.

$\endgroup$
  • $\begingroup$ @dfeuer: Thanks. $\endgroup$ – Ross Millikan Dec 9 '13 at 1:46
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ And the left hand side doesn't even exist for $x\le 15$ ... $\endgroup$ – Henning Makholm Dec 9 '13 at 1:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.