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Is it possible to calculate and find the solution of $ \; \large{105^{1/5}} \; $ without using a calculator? Could someone show me how to do that, please?

Well, when I use a Casio scientific calculator, I get this answer: $105^{1/5}\approx " 2.536517482 "$. With WolframAlpha, I can an even more accurate result.

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    $(105)^{1/5}$ is not square root of $105$.2012-04-02
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    Yes, sorry about that, you're right. I'm editing it right now...2012-04-02
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    I think it's better now, right...? :)2012-04-02
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    Perfect! +1 to the question!2012-04-02
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    @SalechAlhasov: You DID VERY GOOD noticing me, Salech. Indeed. :)2012-04-02
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    Why does this have some many upvotes?2012-04-30
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    @PeterTamaroff: I am not sure but, they wanted it to be so, they might wanted me to keep going about "something nice"...or...they might wanted to encourage me for "being a nice example"... I am really not sure about that... They just let it to be so, they could "punish" me for this otherwise... Why don't you just take it as "a nice example" and 'keep studying Maths on'... :)2012-04-30
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    @KerimAtasoy Good for you then!2012-04-30
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    Thanks for the edit, Antonio.2012-05-03

5 Answers 5

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You can just do it by trial, but it gets tiring: $2^5\lt 105 \lt 3^5$ so it is between $2$ and $3$. You might then try $2.5^5 \approx 98$ so the true value is a bit higher and so on.

An alternate is to use the secant method. If you start with $2^5=32, 3^5=243$, your next guess is $2+\frac {243-105}{243-32}=2.654$ Then $2.654^5=131.68$ and your next guess is $2.654-\frac {131.68-105}{131.68-32}=2.386$ and so on. Also a lot of work.

Added: if you work with RF engineers who are prone to use decibels, you can do this example easily. $105^{0.2}=100^{0.2}\cdot 1.05^{0.2}=10^{0.4}\cdot 1.01=4 dB \cdot 1.01= (3 dB + 1 dB)1.01=2 \cdot 1.25 \cdot 1.01=2.525$, good to $\frac 12$%, where $1.05^{0.2}\approx 1.01$ comes from the binomial $(1+x)^n\approx 1+nx$ for $x \ll 1$

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    Interesting... I'm wondering how do you get these results... :) You guys have very good skills actually... :)2012-04-02
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    @KerimAtasoy: The strategies are well known in numerical analysis. In a sense, I cheated, as I used Wolfram Alpha as a calculator to get the numerics. But I believe they are doable by hand if you are determined enough.2012-04-02
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    @RossMillikan please see my recent question too on mertens theorem !2014-01-02
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I go back to the days BC (before calculators). We did have electricity, but you had to rub a cat's fur to get it.

We also had slide rules, from which a $2$ to $3$ place answer could be found quickly, with no battery to go dead in the middle of an exam. Engineering students wore theirs in a belt holster. Unfortunately, slide rules were expensive, roughly the equivalent of two meals at a very good restaurant. For higher precision work, everyone had a book of tables.

My largish book of tables has the entry $021189$ beside $105$. This means that $\log(105)=2.021189$ (these are logarithms to the base $10$, and of course the user supplies the $2$). Divide by $5$, which is trivial to do in one's head (multiply by $2$, shift the decimal point). We get $0.4042378$.

Now use the tables backwards. The log entry for $2536$ is $404149$, and the entry for $2537$ is $414320$. Note that our target $0.4042378$ is about halfway between these. We conclude that $(105)^{1/5}$ is about $2.5365$.

The table also has entries for "proportional parts," to make interpolation faster. As for using the table backwards, that is not hard. Each page of the $27$ page logarithms section has in a header the range of numbers, and the range of logarithms. The page I used for reverse lookup is headed "Logs $.398\dots$ to $.409\dots$."

There are other parts of the book of tables that deal with logarithms, $81$ pages of logs of trigonometric functions (necessary for navigation, also for astronomy, where one really wants good accuracy). And of course there are natural logarithms, only $17$ pages of these. And exponential and hyperbolic functions, plus a few odds and ends.

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    Very interesting... Glad that you made this far and sharing your experiences world wide... :)2012-04-02
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    ...but are not these numbers coming from answers like the others?!2015-04-21
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    The tables, sure, they depend on the patient work of generations past. I should take this opportunity to mention that good "ballpark" mental estimates were part of the culture in all scientific disciplines. They still are, though I think much less than they were, particularly at the student level.2015-04-21
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You can try using binomial theorem for real exponents.

You can write this as

$$ ((2.5)^5 + (105 - (2.5)^5))^{1/5} = 2.5 \left(1 + \frac{105 - (2.5)^5}{2.5^5}\right)^{1/5} = \frac{5}{2} \left(1 + \frac{47}{625}\right)^{1/5}$$

Taking first three terms of the binomial series

$$(1+x)^r = 1 + rx + \frac{r(r-1)x^2}{2!} + \frac{r(r-1)(r-2)x^3}{3!} + \dots$$

using $r = \frac{1}{5}$ and $x = \frac{47}{625}$ gives us

$$ \frac{5}{2} \left(1 + \frac{47}{5*625} - \frac{4 * 47^2}{2*5^2*625^2}\right) = \frac{4954041}{1953125} \approx 2.5365$$

If you need a better approximation, you can include more terms.

All this can be done by hand using integer arithmetic, but is tedious.

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    VERY GOOD!... :) Thank you very much!... I should better study about these terms and subjects also... :) Thank you again...2012-04-02
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    @KerimAtasoy: You are welcome. But I am curious, what prompted you to ask this question? Are you trying to figure out how calculators work under the covers?2012-04-02
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    :) Sure, it's a nice pleasure to have a such talk with you, sir. Well, I like calculators very much, but calculating and solving some problems without using a calculator seems better sometimes... :) I've already noticed that there're just a few people concern about these approachings in real... :) Yes, I've been also curious about calculators, sometimes, I try to study about them using some scripting or programming languages. But, actually, it's a very hard work to create a very good calculator application, see... :)2012-04-02
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    Notice that using the 1st term of the binomial expansion is similar to a Newton update. To wit, suppose $x$ is an approximation to $105^{\frac{1}{5}}$, then we can write $105^{\frac{1}{5}} = (x^5+(105-x^5))^{\frac{1}{5}} = x(1+\frac{105-x^5}{x^5})^{\frac{1}{5}}$. Expanding the term in parentheses to the first term gives $x(1+\frac{1}{5}\frac{105-x^5}{x^5})$ which simplifies to the Newton update.2012-04-02
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    @KerimAtasoy: Yes, there is a lot of theory involved which can be quite interesting. Good luck with your studies!2012-04-02
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    @user27978: Right! Good observation :-)2012-04-02
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    @Aryabhata: I've been studying about Statistics for a while since I'm having an exam this Weekend, and, I've encountered a problem which needs to apply some "Algebra" and "Roots" in. ( http://en.wikipedia.org/wiki/Geometric_mean ) And, the answer was lying behind this: $ \large{1680^{1/5}} \; $ That's why also. :)2012-04-02
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    @user27978: Thank you very much!... :)2012-04-02
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    How can someone calculate $\frac{4954041}{1953125}$ "without" a calculator? isn't that the objective of this question?2014-08-26
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    @MonK: You can compute it by hand. For instance: http://en.wikipedia.org/wiki/Long_division.2014-08-26
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    Where did you get the $2.5$ number? Is this just a "starting guess" for the root, or is there some other reason for this choice?2014-10-19
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    @anorton: I actually don't remember the reason. But a starting guess for the root is a good guess. It also includes 2 and 5, which typically lead to easier calculations. Also, to use binomial theorem, the guess $g$ you make must satisfy $\frac{|105 - g^5|}{g^5} < 1 $ and the closer to zero that is, the fewer terms you have to compute.2014-10-26
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I'm not exactly sure what you mean by 'without a calculator'.

You could try Newton's method to solve $f(x) = 0$, where $f(x) = x^5-105$. The Newton update is then $x_{n+1} = \frac{4}{5}x_n + \frac{1}{5} \frac{105}{{x_n}^4}$. This converges very quickly.

Of course, this involves computing the 4th power, and dividing...

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    :) Well, How about "... without using a calculator" ..? OK, I'm editing it again... :)2012-04-02
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    I didn't get this method... Could you explain and show it more, please? For example, where that $ \; \large{\frac{4}{5}} \; $ comes from...?2012-04-02
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    Newton's method is based on a linear approximation of a function using the Maclaurin series: $$ f(x)=f(x_0)+f'(x_0)(x-x_0)+O(x-x_0)^2\tag{1} $$ where $O(x-x_0)^2$ is considered insignificant. Trying to find the $x$ so that $f(x)=0$, $(1)$ says $$ x-x_0=-\frac{f(x_0)}{f'(x_0)}+O(x-x_0)^2\tag{2} $$ which, ignoring $O(x-x_0)^2$, leads to the iteration $$ x=x_0-\frac{f(x_0)}{f'(x_0)}\tag{3} $$ Plugging $f(x)=x^5-a$ into $(3)$ yields $$ \begin{align} x &=x_0-\frac{x_0^5-a}{5x_0^4}\\ &=\frac45x_0+\frac{a}{5x_0^4}\tag{4} \end{align} $$2012-04-02
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    :) Thank you very much, robjohn!...2012-04-02
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    My advice if using this and other formulae like it would be to rewrite $x_n$ as $p_n/q_n$, then plug these into the formula to get an expression for $x_{n+1}=p_{n+1}/q_{n+1}$ - you can then divide this into two series for the $p$s and $q$s.2012-04-03
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    In our case we would have $$ \frac{p_{n+1}}{q_{n+1}}=\frac{4}{5}\frac{p_n}{q_n} + \frac{a}{5} \frac{q_n^4}{p_n^4} = \frac{4 p_n^5 + aq_n^5}{5p_n q_n^4} $$ Giving the two series $$ p_{n+1} = 4 p_n^5 + aq_n^5 ,\;\; q_{n+1} = 5p_n q_n^4 $$ Doing this has the advantage that all the subsequent calculations will involve integers (as the $p$s and $q$s are integers) and not messy fractions. You can then long divide at any time to yield a decimal answer2012-04-03
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    I like this rational approach.2012-04-04
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    @JamesFennell and of course robjohn: These are the two most amazing comment entries that I have ever seen. Ever. Anywhere. Wow. Especially robjohn's but then, he has the diamond insignia, so he has special powers ;o) Regardless, this is kind of breathtaking, no hyperbole intended.2012-10-10
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Another way of doing this would be to use logarithm, just like Euler did: $$ 105^{1/5} = \mathrm{e}^{\tfrac{1}{5} \log (105)} = \mathrm{e}^{\tfrac{1}{5} \log (3)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (5)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (7)} $$ Use $$\log(3) = \log\left(\frac{2+1}{2-1}\right) = \log\left(1+\frac{1}{2}\right)-\log\left(1-\frac{1}{2}\right) = \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{2^{2k+1}} = 1 + \frac{1}{12} + \frac{1}{80} + \frac{1}{448} = 1.0.83333+0.0125 + 0.0022 = 1.09803$$ $$ \log(5) = \log\frac{4+1}{4-1} + \log(3) = \log(3) + \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{4^{2k+1}} = \log(3) + \frac{1}{2} + \frac{1}{96} +\frac{1}{2560} $$ $$ \log(7) = \log\frac{8-1}{8+1} + 2 \log(3) = 2 \log(3) - \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{8^{2k+1}} = 2 \cdot \log(3) - \frac{1}{4} - \frac{1}{768} $$ Thus $$ \frac{1}{5} \left( \log(3) + \log(5) + \log(7)\right) = \frac{4}{5} \log(3) + \frac{1}{5} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{96} - \frac{1}{768} + \frac{1}{2560} \right) = \frac{4}{5} \log(3) + \frac{1993}{38400}= 0.9303 = 1-0.0697 $$ Now $$ \exp(0.9303) = \mathrm{e} \cdot \left( 1 - 0.0697 \right) = 2.71828 \cdot 0.9303 = 2.5288 $$

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    :) AMAZING!... I hadn't even noticed about these approachings untill now, sir. Thank you very much!... :)2012-04-02