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The number of non-negative real roots of $2^{x}-x-1$ are

$ a.)\ 0\\ b.) \ 1 \\ c.)\ 2 \\ d.)\ 3 \\$

I don't have any clue.

I have only learned to solve quadratics and cubic equations ,

i haven't been taught to solve such type of equations where

$x$ is written in power.

I look for a short and simple way.

I would prefer a way $\color{red}{\text{without calculus}}$ unless necessary.

I have studied maths up to $12$th grade .Thanks.

R K
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    For us from elsewhere, does "maths up to 12th grade" include differentiation? – Daniel Fischer Jun 21 '15 at 14:47
  • i know basic differetiation and integration . – R K Jun 21 '15 at 14:48
  • Note that by inspection $x=0$ and $x=1$ are non-negative roots. The derivative is $(\ln 2)2^x-1$. Use this to find where the function is decreasing, and where it is increasing. – André Nicolas Jun 21 '15 at 14:52
  • Graph will be a better way – Satvik Mashkaria Jun 21 '15 at 15:07
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    You can graph $y = x + 1$ and $y = 2^{x}$. Where would they intersect. Where would they intersect with x >= 0 ? In how many points ? Does this help ? – Shailesh Jun 21 '15 at 15:27
  • @ shailesh: how did u split $y=2^{x}-x-1$ in two parts $y=x+1$ and $y=2^x$ like that is that legal – R K Jun 21 '15 at 15:29
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    If you are looking for the non-negative roots of 2^x-x-1, that is the same as saying: When does 2^x-x-1=0? By adding x + 1 to both sides you would get 2^x=x+1 The two sides will be equal when the graphs of both sides, 2^x and x+1 intersect. – Amy B Jun 23 '15 at 23:13

4 Answers4

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Consider the function $$f(x)=2^x-x-1$$ Compute the derivative $$f'(x)=2^x \log (2)-1$$ The derivative cancels for $$x_*=-\frac{\log (\log (2))}{\log (2)}$$ For this value $$f(x_*)=-1+\frac{1}{\log (2)}+\frac{\log (\log (2))}{\log (2)}\approx -0.0860713$$ The second derivative test would show that this is a minimum. So, two real roots.

By inspection, $x=0$ is a root and $x=1$ another. These are the roots.

Claude Leibovici
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  • what is the significance of finding $f(x_{*})$ – R K Jun 21 '15 at 15:23
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    This is the minimum value of $f(x)$ and it is negative; so two roots. If it had been positive, then no root at all. Just draw something like that (a parabola would be fine) and it will be totally clear (I hope and wish). Tell me if this works for you. Cheers :-) – Claude Leibovici Jun 21 '15 at 15:26
  • and how did u find $x_{*}$ ? – R K Jun 21 '15 at 15:32
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    $f'(x)=0$ so $2^x=\frac 1 {\log(2)}$ and take logarithms of both sides to isolate $x$. – Claude Leibovici Jun 21 '15 at 15:34
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We can be really fancy about this, but I think there's a very simple solution.

$f(x)=0$ is the same as $2^x=x+1$. The left-hand side is an exponential curve; the right-hand side is a line.

Exponential curves can intersect a line at most twice. If you have a good feel for what exponential curves look like, this will be clear, intuitively; if not, you're going to have to use calculus to prove this (i.e., be fancy).

Luckily for us, there are two readily apparent and easy to see solutions: $f(0)=f(1)=0$.

Since we have found two solutions, and we know there can't be any more, we're done--there are exactly two solutions.

MichaelChirico
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  • Saying that exponential curves intersect a line at most twice is correct, but surely not an answer. To prove it, you would need to resort to those "really fancy" methods that you allude to (and that the OP would prefer not to deal with). I find your answer correct but somehow dishonest, since it hides the garbage under the carpet. – Alex M. Jun 30 '15 at 14:59
  • I understand your hesitation, and I think @wythagoras has the best proof-directed answer; I posted this only given OP's stated aversion to calculus-based answers, and given the question is multiple choice (so an intuitive answer seems valid). I don't know why you think "dishonest", however--I openly state that we'll need calculus if we want this answer to turn into a proof. – MichaelChirico Jun 30 '15 at 15:10
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You could use Bernoulli's inequality, but calculus is used to prove it.

This states

$(1+r)^x \geq 1+rx$ if $x\geq 1$ and $r > -1$ with equality iff $r=0$ or $x=1$.

And $(1+r)^x < 1+rx$ if $0 < x < 1$ and $r > -1$ with equality iff $r=0$.

Rewrite it to $2^x=x+1$.

Insert $r=1$ gives $2^x \geq 1+x$ for $x\geq1$ with equality iff $x=1$.

And $2^x < 1+x$ iff $0 < x < 1$, so no equality here.

We have $x=0$ left, in which case there is equality.

Therefore the only roots are $x=0$ and $x=1$.

Community
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wythagoras
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0

This is a hit and trial method.

You can graph $f(x)=2^x-x-1$

OR

Since you are looking for non-negative real root(s) and you know that $0$ is the smallest non-negative real number so you first find $f(0)$. You will find that $f(0)=0$. Now with intuition decide what next value of $x$ you should take. Take $x=0.5$, now f(x) is less than zero. $f(1)=0$. Now for any value of $x>1$ you will have $f(x)>0$. Therefore there are no more non-negative real solutions.

Remember : Calculus was made to make calculations easy and short as compared to other alternatives.

Warning : What I have done is not a proof. Here the problem was easy and there were options so hit and trial may work. If you are taking any exam which have options then this hit and trial may be used.

If the equation would have been more complicated then I guess that you will love the beauty of calculus because then only you will come to understand why we do calculus.

Singh
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