One Real Root

Given a equation:ax−x=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
7 years, 9 months ago

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Comments

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=logae⇒lna=1e⇒a=e1e\Rightarrow e = log_{a}e \Rightarrow lna = \frac{1}{e} \Rightarrow a = e^{\frac{1}{e}}.

jatin yadav - 7 years, 9 months ago

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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 - 7 years, 9 months ago

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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 - 7 years, 9 months ago

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@Mark Hennings – thanks!

jatin yadav - 7 years, 9 months ago

thank you so much Jatin

idham muqoddas - 7 years, 9 months ago

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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. ex−ax=0 e^x - ax =0

  2. ex−ax2=0 e^x - ax^2 =0

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

jatin yadav - 7 years, 9 months ago

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

jatin yadav - 7 years, 9 months ago

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show me the way, please!

idham muqoddas - 7 years, 9 months ago

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 ln⁡xx  =  ln⁡a \frac{\ln x}{x} \; = \; \ln a The function y=ln⁡xxy = \tfrac{\ln x}{x} increases from −∞-\infty to e−1e^{-1} as xx varies from 00 to ee, and decreases from e−1e^{-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<a≤10 < 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 - 7 years, 9 months ago

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 ax∗lna=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=logae⇒a1lna=e=1lna→lna=1e⇒a=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 - 7 years, 9 months ago
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