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# C2 Exponential Graph and Log question

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1. I am really stuck on 5c and i have got the mark scheme but I dont understand it . any help would be appreciated
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2. (Original post by D_Breezy)
I am really stuck on 5c and i have got the mark scheme but I dont understand it . any help would be appreciated
Which bit don't you undertstand?

We can write , so if we equate the two equations we have

Multiply both sides by the denominator of the LHS:

but you know what the same thing multiplied by itself is, don't you? It's a square. So:

. Now divide both sides by 2:

. Now take the square root:

. reject the negative value because exponentials are always positive.

SO . Take logarithms to solve: . Isolate .
3. (Original post by Zacken)
Which bit don't you undertstand?

We can write , so if we equate the two equations we have

Multiply both sides by the denominator of the LHS:

but you know what the same thing multiplied by itself is, don't you? It's a square. So:

. Now divide both sides by 2:

. Now take the square root:

. reject the negative value because exponentials are always positive.

SO . Take logarithms to solve: . Isolate .
hey thanks i just forgot
. thank you !
4. (Original post by D_Breezy)
hey thanks i just forgot
. thank you !
Yep, just remember that and that in this case, since one to the power of anything is 1.
5. （1/3）^x =2（3）^x
1/3=3^（-1）
thus （3）^（-x）=2（3）^x
multiplied by （3）^x on both sides
then 1= 2（3）^（2x）
（3）^（2x）=1/2
2x=lg（1/2）/lg（3）
x=1/2lg（1/2）/lg（3）
using the calculator，x=-0.32
then substitute x= 1/2lg（1/2）/lg（3）into y=（1/3）^x，1/2lg（1/2）/lg（3）=1/2log3（1/2）=log3（square root of 1/2）
then y= （1/3）^x= 3^（-x）=3^（-log3（square root of 1/2）=3^log3（square root of 2）=root 2

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6. what exam board do you do?
7. Looks like an Edexcel question.

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