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Hello i must find t in $1500=1500(1-e^{t/0,54})$

I tried $1=1-e^{t/0,54}$ Then $0=e^{t/0,54}$ But i don't know what to do here cause i can't use the logarithm

Thanks for your answers

UdWeed
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    The equation is impossible as the exponential can never be $0$. – lulu Dec 19 '17 at 21:08
  • By showing that $0 = e^{\frac t{0.54}}$ you have proven that there is no possible solution. That is an acceptable final result. If you were to graph this you would see it is assymptotic on the negative/left side to 1500. All possible values are less than 1500, with large enough negative values being as arbtririly close to 1500 as we like but never actually equaling 1500. And you HAVE shown that. – fleablood Dec 19 '17 at 22:22

2 Answers2

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Maybe not exactly an answer but the MathJax somehow doesn't render in comment on my device.

I think you maybe looking at the situation where $$\lim_{t\to-\infty}\left[{1500\left(1-e^{t/0.54}\right)}\right]=1500(1-0)=1500$$

So the $t$ you are looking for is probably $t\to-\infty$. Not sure what's the context of this. It probably just translates to "the model has value $1500$ at the very start of time, and value $0$ at the point of first observation."

Karn Watcharasupat
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  • " doesn't render in comment on my device" Hear, hear! That drives me NUTS! – fleablood Dec 19 '17 at 22:23
  • @fleablood sorry I don't really get what you are trying to say... – Karn Watcharasupat Dec 19 '17 at 22:26
  • I'm sympathizing with the Phone App of StackExchange not rendering MathJax in comments. It drives me nuts. The four things that drive me nuts about the phone app in order are: 1) You can't view the question as you write an answer. 2) Comments do not render MathJax 3) you can not delete answers 4) You can either cut and paste an entire comment or not cut and paste at all. – fleablood Dec 19 '17 at 22:49
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$$ 1500=1500\left(1-e^{\frac{t}{0.54}}\right)\implies\\ 1=1\cdot\left(1-e^{\frac{t}{0.54}}\right)\implies\\ 1=1-e^{\frac{t}{0.54}}\implies\\ e^{\frac{t}{0.54}}=0 $$

By the very definition of the exponential function ($f(x)=a^x$ where $a>0$ and $a≠1$), there is no positive number raised to a power that's equal to zero. In other words, you can try raising a number to any power all you want, you will never be able to get it equal to zero. So, your equation has no solutions.

Michael Rybkin
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