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# what do i do to find k

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1. The equation f(t)=ke^t, where k>0, has one real solution.
Find the value of k.

I tried working it out and couldn't do it. Then I plotted it on a graph and saw that the only time there was a root (as far as I could tell), was when k=0 which had infinite roots as well as k not being positive.
2. (Original post by KloppOClock)
The equation f(t)=ke^t, where k>0, has one real solution.
Find the value of k.

I tried working it out and couldn't do it. Then I plotted it on a graph and saw that the only time there was a root (as far as I could tell), was when k=0 which had infinite roots as well as k not being positive.
As it stands you just have a function. You need to equate it to something to get an equation that can be solved. You need to check the exact wording of your original question. Or post a link/photo.
3. (Original post by ghostwalker)
As it stands you just have a function. You need to equate it to something to get an equation that can be solved. You need to check the exact wording of your original question. Or post a link/photo.
4. (Original post by KloppOClock)
This is definitely the whole question? There aren't any earlier parts where f(t) is defined? Because that question doesn't make any sense- you need more information.
5. (Original post by sindyscape62)
This is definitely the whole question? There aren't any earlier parts where f(t) is defined? Because that question doesn't make any sense- you need more information.
6. (Original post by KloppOClock)
Ok, now it's solvable- do you see why? Part c puts f(t) equal to something else, so if you don't know what f(t) is (defined at the start) you can't get anywhere with it.

Your first step is to write
Then solve it for t. Try multiplying through by and you should get a quadratic. Quadratics either produce 0, 1 or 2 different real roots- think about the condition for a quadratic to produce just one. This will give you a solvable equation in terms of k.
7. (Original post by sindyscape62)
Ok, now it's solvable- do you see why? Part c puts f(t) equal to something else, so if you don't know what f(t) is (defined at the start) you can't get anywhere with it.

Your first step is to write
Then solve it for t. Try multiplying through by and you should get a quadratic. Quadratics either produce 0, 1 or 2 different real roots- think about the condition for a quadratic to produce just one. This will give you a solvable equation in terms of k.
oh right i forgot about the original fx bit, thanks

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