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1. 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
2. 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
3. i) replace ex with a different letter, W say....

3W + 28/W = 25
4. (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!
5. (Original post by 13 1 20 8 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?
6. (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:
Show
You can do e^(LHS) = e^(RHS).
7. (Original post by 13 1 20 8 42)
Well think about what the natural log fundamentally means and it will be simple. If not,
Spoiler:
Show
You can do e^(LHS) = e^(RHS).
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?
8. (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.
9. (Original post by 13 1 20 8 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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