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#1
Have no idea how to approach this question g
e^x-4= 2e^2x

Posted from TSR Mobile
0
5 years ago
#2
(Original post by Hardworker?)
Have no idea how to approach this question g
e^x-4= 2e^2x

Posted from TSR Mobile
A quadratic in disguise (he said, risking the wrath of TenOfThem)
2
5 years ago
#3
If it's e^(x - 4), take logs of both sides, bearing in mind ln(ab) = ln(a) + ln(b).
If it's , then use e^x = y. Which will turn the equation into:

y - 4 = 2y^2
0
5 years ago
#4
(Original post by Hardworker?)
Have no idea how to approach this question g
e^x-4= 2e^2x

Posted from TSR Mobile
Hadn't occurred to me, but from SecretDuck's posting, this may be open to interpretation.

Do you mean:

, a la Duck.

Or do you mean

, moi.

0
5 years ago
#5
(Original post by SecretDuck)
Take logs of both sides, bearing in mind ln(ab) = ln(a) + ln(b)
That's not going to help if the LHS is
0
5 years ago
#6
(Original post by davros)
That's not going to help if the LHS is
Ghostwalker and I interpreted the question in different ways. If it's , then you turn it into a quadratic.

I edited that post to take both cases into account
0
5 years ago
#7
(Original post by SecretDuck)
Ghostwalker and I interpreted the question in different ways. If it's , then you turn it into a quadratic.
Yeah, the OP needs to use brackets or Latex - these are always a pain to decipher!!
0
5 years ago
#8
(Original post by davros)
Yeah, the OP needs to use brackets or Latex - these are always a pain to decipher!!
I posted both interpretations regardless
0
5 years ago
#9
(Original post by SecretDuck)
I posted both interpretations regardless
We are of course ignoring the possibility , but as that has no analytical solution, it's fairly safe, unless it's a numerical methods module. Ho, hum.
0
5 years ago
#10
(Original post by ghostwalker)
We are of course ignoring the possibility , but as that has no analytical solution, it's fairly safe, unless it's a numerical methods module. Ho, hum.
Well, I did attempt both interpretations and I got imaginary numbers for some reason. Not sure if I'm doing anything wrong
0
5 years ago
#11
(Original post by SecretDuck)
Well, I did attempt both interpretations and I got imaginary numbers for some reason. Not sure if I'm doing anything wrong
I get the same, with my interpretation, though yours doesn't require complex x.

Over to the OP, as they say.

Edit: Correction.
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