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We know that for any value of $x$ other than $0$, $a^x\ne e^x$ where $a>e, a\in R^+$ but we do know that for some value of $p$, $$pa^x=e^x\ldots(1)$$

you see $p$ is a positive number because of the above assumption. we all know that $$\frac{\operatorname{d}e^x}{dx}=e^x, \frac{\operatorname{d}a^x}{dx}=a^x\times \log_e a$$

so using above known things, i am going to differentiate equation $(1)$ with respect to $x$

$$\frac{\operatorname d \left(pa^x\right)}{dx}=\frac{\operatorname de^x}{dx}\implies pa^x\ln a = e^x\ldots(2)$$ Note that $p$ is a number and i am talking at a particular value of $x$.

Now from (1) and (2) we can say that $$pa^x=pa^x\ln a\implies a=e,$$

but earlier it was taken that $a>e$ but this turn out to be $a=e$.

But you see at a particular value of $x$ which could be any number, $a$ will comes equal to $e$.

So, what should i conclude from the above?

Davide Giraudo
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Aster rovels
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    It's good to learn from mistakes. As Andrew's answer points out, just because two functions $f(x)=g(x)$ at some isolated value of $x$ does not imply $f'(x)=g'(x)$. So everything after you try to differentiate (1) is mistaken. You might try solving (1) for $p$ to see that for fixed $a$, the value of $p$ depends on $x$, so that correctly differentiating (1) with repsect to $x$ requires us to account for that dependence. – hardmath Jun 30 '16 at 12:18

1 Answers1

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When you write $$ pa^{x} = e^{x}, \tag{1} $$ you're saying (1) holds for one single $x$. When you differentiate to obtain $$ pa^{x} \ln a = e^{x}, \tag{2} $$ you're implicitly assuming (1) holds for all $x$ in some open interval.

That is, (2) does not follow from (1) in your situation; any conclusion obtained jointly from (1) and (2) is logically suspect. (Compare Differentiating both sides of a non-differential equation.)

Community
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Andrew D. Hwang
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