# One Real Root

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

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

Note by Idham Muqoddas
5 years, 1 month ago

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## Comments

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Consider two curves $$y = a^x$$ and $$y = x$$. 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, $$a^x lna = 1$$ $$\Rightarrow$$ $$a^x = \frac{1}{lna}$$ . Since $$a^x = x$$ $$\Rightarrow$$ $$x = \frac{1}{lna}$$. Hence we get that $$a^{\frac{1}{lna}} = \frac{1}{lna}$$. Now $$\frac{1}{lna} = log_{a}e$$ . Hence LHS becomes e. $$\Rightarrow e = log_{a}e \Rightarrow lna = \frac{1}{e} \Rightarrow a = e^{\frac{1}{e}}$$.

- 5 years, 1 month ago

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thank you so much Jatin

- 5 years, 1 month 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. $$e^x - ax =0$$

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

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

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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

- 5 years, 1 month 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!

- 5 years, 1 month ago

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thanks!

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

Substituting $$a^x = x$$ we get $$x = \frac {1}{ln a}$$. Now, substituting this back in to the original equation we get $$a^{\frac {1}{ln a}} = \frac {1}{ln a} = \frac {ln e}{ln a} = log_ae \Rightarrow a^{\frac {1}{ln a}} = e = \frac{1}{ln a} \rightarrow ln a = \frac {1}{e} \Rightarrow a = e^{\frac {1}{e}}$$

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

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

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

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

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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show me the way, please!

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