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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. $\Rightarrow e = log_{a}e \Rightarrow lna = \frac{1}{e} \Rightarrow a = e^{\frac{1}{e}}$.

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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 $a > e^{\frac{1}{e}}$.

$y = x$ 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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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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## Comments

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TopNewestConsider 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. $\Rightarrow e = log_{a}e \Rightarrow lna = \frac{1}{e} \Rightarrow a = e^{\frac{1}{e}}$.

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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

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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!

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thank you so much Jatin

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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.

$e^x - ax =0$

$e^x - ax^2 =0$

$x = ln(ax)$ , etc.

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

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show me the way, please!

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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 $a > e^{\frac{1}{e}}$.

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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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