1
$\begingroup$

$$2^{x-3}=\frac{1}{x}$$

So far, I've only managed to solve it graphically. I was wondering if there is any other method available? I know about the $\ln$ method of course.

$\endgroup$
  • $\begingroup$ Observation is a way $\endgroup$ – randomgirl Jun 27 '17 at 22:25
  • $\begingroup$ @randomgirl I was thinking the same thing but it seemed quite non-mathematical approach. But hey :) $\endgroup$ – kenobe Jun 27 '17 at 22:26
2
$\begingroup$

A non-iterative approach is to use Lambert's W function.

$$\begin{align} &2^{x-3}=\frac{1}{x}\\ \implies &x2^x=8 \\ \implies&xe^{x\ln 2}=8 \\ \implies&x\ln2\cdot e^{x\ln 2}=8\ln2 \\ \implies&x=\frac{W(8\ln2)}{\ln2} \\ \text{(Thanks}&\text{ to projectilemotion for the following:)} \\ \implies&x=\frac{W(4\ln4)}{\ln2} \text{ (Using identity: } W(x\ln x) = \ln x)\\ \implies&x=\frac{\ln4}{\ln2} = 2\\ \end{align}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Wow so many methods. Thanks $\endgroup$ – kenobe Jun 28 '17 at 0:17
  • 1
    $\begingroup$ We can simplify this further: $$\frac{W(8\ln{2})}{\ln{2}}=\frac{W(4\ln{4})}{\ln{2}}=\frac{\ln{4}}{\ln{2}}=\frac{2\ln{2}}{\ln{2}}=2$$ Where we used the identity $W(x\cdot \ln{x})=\ln{x}$ for $x>0$. $\endgroup$ – projectilemotion Jun 28 '17 at 20:00
  • $\begingroup$ @projectilemotion Thanks. Good addition. $\endgroup$ – Χpẘ Jun 28 '17 at 20:02
5
$\begingroup$

Since the left hand side is positive, $x$ must be positive.

The problem is equivalent to $$x2^x=8$$ Notice that product of positive increasing function is increasing and hence it has a unique solution.

$$x2^x=2^3=2(2^2)$$

Hence $2$ is the unique solution.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Do you mean strictly increasing in the interval $] \infty, \infty [$? $\endgroup$ – Toby Mak Jun 27 '17 at 22:45
  • $\begingroup$ I like your approach. Thanks ! You are awesome $\endgroup$ – kenobe Jun 27 '17 at 22:46
  • $\begingroup$ I mean for $(0, \infty)$ $\endgroup$ – Siong Thye Goh Jun 27 '17 at 22:46
1
$\begingroup$

The other obvious method is bisection -- find a value for which the left side is less than the right, a value for which it's greater, and then check the midpoint. Repeat.

Of course, a first step is to guesstimate and try some integers; I did so and found $x = 2$ on my third try (after I'd checked $x = 1, 3$ becasue they were algebraically really easy).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I like the bisection. Thank you very much sir. $\endgroup$ – kenobe Jun 27 '17 at 22:29
  • $\begingroup$ My solution can be regarded as the completion of @SiongThyeGogh's, for while his tells you that there is a unique solution, mine gives you a very practical (for this particular problem) way to approximate that solution to very high accuracy. $\endgroup$ – John Hughes Jun 27 '17 at 23:47
  • $\begingroup$ Warning: Only works for intervals where the given function is monotone. $\endgroup$ – Simply Beautiful Art Jun 27 '17 at 23:52
  • 1
    $\begingroup$ No. It only works where the function is continuous and you can find an interval on which the function has opposite signs at the two ends. When you evaluate at the midpoint, the value is either $0$ (yay!), the opposite sign from the left end, or the opposite sign from the right end. Whichever half-interval has opposite-sign ends contains a root. Iterate on that interval, thereby constraining some root in exponentially-nested intervals, as claimed. Regardless, the other answer showed that it's continuous and increasing, so we're OK in either case. $\endgroup$ – John Hughes Jun 27 '17 at 23:56
  • $\begingroup$ It is true that the root is guaranteed to be unique only if the function is monotone, unless you have some other information about the function. $\endgroup$ – John Hughes Jun 28 '17 at 0:16

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.