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I have to find roots of equation $x3^x=1$

A.Infinitely many roots

B.$2$ roots

C.$1$ root

D. No roots\

How do i start? Thanks

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4 Answers 4

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Hint:

$$3^x = \frac{1}{x}$$ Has only $1$ solution.

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    $\begingroup$ @SophieClad: how can you accept this solution which gives absolutely no justification ? $\endgroup$
    – user65203
    Jan 17, 2016 at 11:58
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    $\begingroup$ It is a multiple choice question and therefore most likely is meant to be solved under time pressure. The hint I gave allows OP to solve the question by using some intuition rather than a complete answer. $\endgroup$
    – fosho
    Jan 17, 2016 at 12:00
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    $\begingroup$ @YvesDaoust i accepted it because i got idea of making graphs of both sides. $\endgroup$ Jan 17, 2016 at 12:01
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    $\begingroup$ @YvesDaoust yes you are right. But i have to sit in exam where i have to tick correct answers. I do not need such deep explanations. $\endgroup$ Jan 17, 2016 at 12:07
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    $\begingroup$ I also upvoted this answer, because A) it gets the job done (the actual question is about the number of solutions), B) I somewhat dismiss Lambert W, and oppose any "political" motion of getting it included in the list of elementary functions. At least until it begins to have so many applications that it is included in standard pocket calculators and such. Mind you, I won't downvote questions/answers using W, I just ignore them. $\endgroup$ Jan 17, 2016 at 12:12
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For a simple answer plot the graphs of $y_1=3^x$ and $y_2=\frac{1}{x}$, (that are elementary), and see that these graph intersects only at a point $x_0$ such that $0<x_0<1$ because:

1) $3^0=1$ and $ \frac{1}{x} \to +\infty$ for $x \to 0^+$

2) $y_1(1)=3^1=3>y_2(1)=\frac{1}{1}=1$

3) the two functions are continuous in $(0,1]$.

4) for $x>0$ $y_1$ is monotonic increasing and $y_2$ is monotonic decreasing and for $x<0$: $y_1>0$ and $y_2<0$.

If you want the value of $x_0$ this cannot be done with elementary functions. You can use the Lambert $W$ function that is defined as: $$ W(xe^x)=x $$ so, from $$ x3^x=1 \iff xe^{x\ln 3}=1 $$ using $x\ln 3=t$ we find: $$ te^t=\ln 3\quad \Rightarrow \quad t=W(\ln 3) $$ and $$ x_0=\frac{W(\ln 3)}{\ln 3} $$

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Define the function $f(x)= x3^x-1= x\exp(ln(3)x)-1$ and study the variation of this function on $\mathbb{R}$.

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We can solve this by Newton raphson method.

$x3^x=1$

By solving i got $x$ approximately equal to 0.5478

The given problem have only one solution( by graphical method )

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