# Exponential Problem

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#1
Hi! Can someone please help me with this question I've attached. I don't understand how I can prove b to = 20 and the mark scheme is confusing me even more. Many thanks 0
4 months ago
#2
what have you done so far?
0
#3
(Original post by Bexr)
what have you done so far?
I got 290 = a + b
and 331 = ae^0.1 +be^0.2

then ae^0.1 = be^0.2 /290
0
4 months ago
#4
Yep - good! So .... rather than rearrange the 2nd equation, think about the b value you are trying to prove and how you can rewrite the second equation with just b. (HINT look at the first "simple" equation)
0
4 months ago
#5
(Original post by Hollymae764)
I got 290 = a + b
and 331 = ae^0.1 +be^0.2

then ae^0.1 = be^0.2 /290
The first two equations are correct, but the third one is wrong.
I would suggest multiplying the first equation (all terms) by e^0.1, then subtracting one equation from the other to eliminate a and leave b.
1
4 months ago
#6
I was trying to get you to spot that if you rewrite the first equation as
a = 290-b
you can then sub this into the 2nd equation, which will give eliminate "a" and allow you to rearrange to find "b".

In any case, by creating the 2 equations you now have simultaneous equations, so you can use your favourite method to solve them.
0
#7
(Original post by Bexr)
I was trying to get you to spot that if you rewrite the first equation as
a = 290-b
you can then sub this into the 2nd equation, which will give eliminate "a" and allow you to rearrange to find "b".

In any case, by creating the 2 equations you now have simultaneous equations, so you can use your favourite method to solve them.
Thank you! That helped so much 0
#8
(Original post by Hollymae764)
Thank you! That helped so much I'm kind of stuck again, so I rearranged them

ae^0.1t = be^0.2t -290

then I put this into the other equation

331 = (be^0.2t -290) + be^0.2t
621 = 2be^0.2t

I don't know what to do from here is this even correct?
0
4 months ago
#9
(Original post by Hollymae764)
I'm kind of stuck again, so I rearranged them

ae^0.1t = be^0.2t -290

then I put this into the other equation

331 = (be^0.2t -290) + be^0.2t
621 = 2be^0.2t

I don't know what to do from here is this even correct?
You seem to have got a bit muddled with the algebra!

Remember that your second equation (for 2018) comes from when t = 1 so you shouldn't have any t's left in your final equations.

You found that a + b = 290 so that a = 290 - b. What do you get first of all when you substitute this into 0
#10
I did:

331 = 290 - be^0.1 + be^0.2
41 = -be^0.1 +be^0.2
41 = -b (e^0.1 - e^0.2)
41/(e^0.1 - e^0.2) = -b
b = 352.74

What have I done wrong?
0
4 months ago
#11
If should be:

331 = (290-b)e^0.1 + be^0.2

(You've subbed in "a" without using brackets)
0
4 months ago
#12
(Original post by Hollymae764)
I did:

331 = 290 - be^0.1 + be^0.2
41 = -be^0.1 +be^0.2
41 = -b (e^0.1 - e^0.2)
41/(e^0.1 - e^0.2) = -b
b = 352.74

What have I done wrong?
When you multiply the equation 290 = a + b by e^0.1, you have to multiply all the terms in the equation, including the 290. Then 290e^0.1 = ae^0.1 + be^0.1.
0
4 months ago
#13
The way that old_engineer suggests is the easiest method - multiply the first simple equation by e^0.1 and then subtract one from the other to eliminate "a"
0
#14
(Original post by mathstutor24)
If should be:

331 = (290-b)e^0.1 + be^0.2

(You've subbed in "a" without using brackets)
ohhh that makes sense
0
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