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Core 3 Maths, e and ln Problems, Please Help

Hi, I have come across 2 types of questions involving e and ln that I can't solve, any one who can solve for the answer I would appreciate your help if you could show me your working out and answer:

1. 3eˣ + 28e⁻ˣ = 25

2. ln(3x+1) - ln(2x-1) = 1

3. ln(2x-1) - ln(3x+1) = 1

4. ln(2x-1) - ln(3x+1) = 0

Thank you
Reply 1
Avatar for math42
math42
20
For the first one you can turn it into a quadratic in terms/powers of e^x. Rewrite the negative power in reciprocal form, then get rid of the fraction by multiplying out and it should be more obvious.

The others you need to use a certain log rule. ln(a) - ln(b) = ln(a/b).

Original post by Clankdroid
x
(edited 8 years ago)
i) replace ex with a different letter, W say....

3W + 28/W = 25
Reply 3
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Clankdroid
OP
1
Original post by the bear
i) replace ex with a different letter, W say....

3W + 28/W = 25


Thank you I see how to make it a quadratic now!
Reply 4
Avatar for Clankdroid
Clankdroid
OP
1
Original post by 1 8 13 20 42
For the first one you can turn it into a quadratic in terms/powers of e^x. Rewrite the negative power in reciprocal form, then get rid of the fraction by multiplying out and it should be more obvious.

The others you need to use a certain log rule. ln(a) - ln(b) = ln(a/b).


After using that log rule tho I don't know how to further solve for x, can you do it please?
Reply 5
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math42
20
Original post by Clankdroid
After using that log rule tho I don't know how to further solve for x, can you do it please?


Well think about what the natural log fundamentally means and it will be simple. If not,

Spoiler

Reply 6
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Clankdroid
OP
1
Original post by 1 8 13 20 42
Well think about what the natural log fundamentally means and it will be simple. If not,

Spoiler



okay, so ln(3x+1) - ln(2x-1) = 1

raise to e, which would make 3x+1 -2x+1 = e

3x-2x= e-2
x=e-2?

but i thought,

ln(3x+1) - ln(2x-1) = 1lne
ln(3x+1/2x-1)=lne
3x+1/2x-1 = e? but then how do you solve, and how about for one where it equaled 0?
Reply 7
Avatar for math42
math42
20
Original post by Clankdroid
ln(3x+1) - ln(2x-1) = 1lne
ln(3x+1/2x-1)=lne
3x+1/2x-1 = e? but then how do you solve, and how about for one where it equaled 0?


That way is right. You would just multiply both sides by (2x - 1), get the terms containing x to one side and the remaining terms to the other side, factorise so you get x on its own and divide to obtain x.
0 = ln1.
Reply 8
Avatar for Clankdroid
Clankdroid
OP
1
Original post by 1 8 13 20 42
That way is right. You would just multiply both sides by (2x - 1), get the terms containing x to one side and the remaining terms to the other side, factorise so you get x on its own and divide to obtain x.
0 = ln1.


thanks for all of your help!

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