One Real Root

Given a equation:axx=0a^x-x=0 for a>1a>1

Determine value of aa so that the equation has exactly 1 real root

Note by Idham Muqoddas
5 years, 11 months ago

No vote yet
4 votes

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Consider two curves y=ax y = a^x and y=x y = x. Since , the two curves would intersect at only one point , we would better like to say that the two curve would touch each other. Or they have same derivative at the point of contact, taking derivative of both, axlna=1 a^x lna = 1 \Rightarrow ax=1lna a^x = \frac{1}{lna} . Since ax=xa^x = x \Rightarrow x=1lna x = \frac{1}{lna} . Hence we get that a1lna=1lnaa^{\frac{1}{lna}} = \frac{1}{lna}. Now 1lna=logae\frac{1}{lna} = log_{a}e . Hence LHS becomes e. e=logaelna=1ea=e1e\Rightarrow e = log_{a}e \Rightarrow lna = \frac{1}{e} \Rightarrow a = e^{\frac{1}{e}}.

jatin yadav - 5 years, 11 months ago

Log in to reply

I request someone to tell me how to break the lines in formatting, the above lines look together but weren't written together

jatin yadav - 5 years, 11 months ago

Log in to reply

If you want to separate paragraphs, you need to leave at least one clear blank line between them. Press the carriage return twice, not once, at the end of each paragraph!

Mark Hennings - 5 years, 11 months ago

Log in to reply

@Mark Hennings thanks!

jatin yadav - 5 years, 11 months ago

Log in to reply

thank you so much Jatin

idham muqoddas - 5 years, 11 months ago

Log in to reply

Now , this method can be extended to a lot of problems .

For example , find the individual values of a so that the given equations have only one solution.

  1. exax=0 e^x - ax =0

  2. exax2=0 e^x - ax^2 =0

  3. x=ln(ax)x = ln(ax) , etc.

jatin yadav - 5 years, 11 months ago

Log in to reply

a=e1e a = e^{\frac{1}{e}}

jatin yadav - 5 years, 11 months ago

Log in to reply

show me the way, please!

idham muqoddas - 5 years, 11 months ago

Log in to reply

Since ax>0a^x > 0 for all real xx, we deduce that the only possible solutions of this equation must be positive. Thus we can restrict attention to positive xx, and then this equation is equivalent to lnxx  =  lna \frac{\ln x}{x} \; = \; \ln a The function y=lnxxy = \tfrac{\ln x}{x} increases from -\infty to e1e^{-1} as xx varies from 00 to ee, and decreases from e1e^{-1} to 00 as xx increases from ee to \infty. Thus

  • The equation has a unique positive solution lying in (0,1](0,1] for any 0<a10 < a \le 1, but we are not interested in this case.

  • The equation has two positive real solutions lying in (1,)(1,\infty) for any 1<a<e1e1 < a < e^{\frac{1}{e}}.

  • The equation has a unique real solution x=ex=e for a=e1ea = e^{\frac{1}{e}}.

  • The equation has no real solutions for a>e1ea > e^{\frac{1}{e}}.

Mark Hennings - 5 years, 11 months ago

Log in to reply

y=xy = x is a line with slope 1 passing through the origin. y=axy = a^x is a curve which contains the point (0,1)(0, 1) and grows exponentially to the right of this point and approaches y=0y = 0 to the left of this point for a>1a > 1. If there is only one solution for xx, then necessarily these two graphs intersect only once, meaning that they have the same derivative at their intersection. Taking derivatives of both we get axlna=1a^x * ln a = 1 .

Substituting ax=xa^x = x we get x=1lnax = \frac {1}{ln a}. Now, substituting this back in to the original equation we get a1lna=1lna=lnelna=logaea1lna=e=1lnalna=1ea=e1ea^{\frac {1}{ln a}} = \frac {1}{ln a} = \frac {ln e}{ln a} = log_ae \Rightarrow a^{\frac {1}{ln a}} = e = \frac{1}{ln a} \rightarrow ln a = \frac {1}{e} \Rightarrow a = e^{\frac {1}{e}}

Michael Tong - 5 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...