# Find exponential function f(x)=Ca^x

linhthuy256
help me to solve this problem
find exponential function f(x)=Ca^x with the 2 given points (0,5)(2,5/9)

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Gold Member
linhthuy256 said:
find exponential function f(x)=Ca^x with the 2 given points (0,5)(2,5/9)
thankzzzzzzzzzz...
Alright, $f(x)=Ca^x$, f(0)=5, and f(2)=5/9. So you have two equations and two unkowns: a and C. You can write the first equation as:
$$5=Ca^0$$
What is a number to the zero power? Answer that and you have C. Now the second equation is:
$$\frac{5}{9}=Ca^2$$
Remember, you already found C from the last equation, so you just have to find a. Just isolate "a" and take the square root of both sides:
$$a=\sqrt{\frac{5}{9C}}$$
Then just plug "a" and "C" back into the original equation and you're done.

linhthuy256
Thanks a lot for ur help. Actually i did the problem 5 times already and i knew C=5 but i can not find a=? and My assignment is about to be due in half an hour. Again thanks for ur help

Gold Member
Have you found "a" now?

linhthuy256

Gold Member
You mean $a=\frac{1}{3}$ right?

linhthuy256
yes and the function f(x)=5*1/3^x that's what i had got before but when i put down in the answer and it said "wrong", can u help me ??

Gold Member
That seems right. Try 5*(1/3)^x, with the parentheses, maybe.

linhthuy256
linhthuy256 said:
yes and the function f(x)=5*1/3^x that's what i had got before but when i put down in the answer and it said "wrong", can u help me ??

beside i still have 1 more problem that is really need help will u able to help me please?

linhthuy256
linhthuy256 said:
beside i still have 1 more problem that is really need help will u able to help me please?

yes i did every single posible way but it stills said "wrong" or the answer is not accepted.

Gold Member
linhthuy256 said:
beside i still have 1 more problem that is really need help will u able to help me please?
Alright, what's the other problem?

Gold Member
linhthuy256 said:
yes i did every single posible way but it stills said "wrong" or the answer is not accepted.
What was the exact phrasing of the problem?

linhthuy256
the exact phrasing was what i posted it on at first.

linhthuy256
and the other problem is
compare the functions f(x)=x^5 and g(x)=5^x by graphing both f and g in several viewing rectangles.
a) find the x-coordinates of the point intersectionof the two curves accurate to two decimal places. and i did the graphing w/ a result of (1.70,1.70) is that rite?
and part b) find the value of x greater than 5 for which g(x)=400 f(x)

linhthuy256
yeah i got the first problem rite. thanks

Gold Member
linhthuy256 said:
and the other problem is
compare the functions f(x)=x^5 and g(x)=5^x by graphing both f and g in several viewing rectangles.
a) find the x-coordinates of the point intersectionof the two curves accurate to two decimal places. and i did the graphing w/ a result of (1.70,1.70) is that rite?
and part b) find the value of x greater than 5 for which g(x)=400 f(x)
No, that's not right, in fact niether function passes through that point. You could graph the two functions in a different window to find the true intersection, but it is clear by inspection that x=5 will yield 5^5 in both cases. As for the second case, just graph 400x^5 instead of x^5 and see where this intersects g(x).

linhthuy256
i can not find the intersection of 400x^5

Gold Member
Look between x=10 and x=12. y will be between 10^6 and 10^8.

linhthuy256
LeonhardEuler said:
Look between x=10 and x=12. y will be between 10^6 and 10^8.

it's still have no intersection point between 10 or 12-13

Gold Member
The intersection point is between x=10 and x=12. Just make sure you're looking at y values around between 10^6 and 10^8.

linhthuy256
huhmm i can find any answers for the intersection. i think i have to leave out this question. do u know any other ways to graph and view the intersection beside do the second trace then press intersection button?

Gold Member
Yes. Go to the "range" window. Put xmin as 10, xmax as 12, ymin as 10^6, and ymax as 10^8. Then just use the intersection feature. Alternitivley, just keep guessing values for x. g(x) starts out smaller than f(x) and then gets bigger. So, if you put in a certain x value and g(x) is smaller than 400x^5, try a bigger x. If g(x) is to big, try a smaller x.

linhthuy256
LeonhardEuler said:
Yes. Go to the "range" window. Put xmin as 10, xmax as 12, ymin as 10^6, and ymax as 10^8. Then just use the intersection feature. Alternitivley, just keep guessing values for x. g(x) starts out smaller than f(x) and then gets bigger. So, if you put in a certain x value and g(x) is smaller than 400x^5, try a bigger x. If g(x) is to big, try a smaller x.

i give up i did as wat u say for both first and second curve but still it doesn't show any intersection point when i go between x=1 and up to x=25

Gold Member
Did you graph f(x)=400x^5 and g(x)=5^x? I don't see what could possibly go wrong. I've found the intersection and checked that it was correct and it was between 10 and 12. Just try guess and test if nothing else is working. But make sure you put everyting in right first.

I have to go to sleep now, so I hope you manage to figure this out now. If not, hopefully someone else will come. Good luck!

linhthuy256
ok i found 1 point that's 11.17 rite?

linhthuy256
thanks a lot and nite nite