Solving for x - exponential equations to base e (1 Viewer)

[Symphonicity]

I have been struggling with this problem all day.

Can anyone walk me through the steps?

solve for x: e^x - 2e^-x = 1.

Many thanks. I find this topic really tough.

 

[Carrotsticks]

Multiply both sides by e^x and re-arrange.

 

[Symphonicity]

I've never done a question like this before, but your answer makes sense. Thanks very much. I may (almost certainly) will be back with more questions. Cheers!

 

[Symphonicity]

Alright, I give up. What about ln(lnx) = 1?

 

[HSC2014]

I don't know how to explain this well but,
ln (ln x) = 1 simply means e^1 = (ln x) - from log definitions (kinda)
and furthermore, x = e^e

 

[Sy123]

Alright, I give up. What about ln(lnx) = 1?

just do what you normally do when solving for a



What would you do normally? raise both sides by e



Similarly:





and then.......

 

[HSC2014]

just do what you normally do when solving for a



What would you do normally? raise both sides by e



Similarly:





and then.......

What does that mean? I see it a lot, but I fail to see the algebra manipulation. I learnt to read/comprehend "ln a = b" as "e to power b equals a" - there is no use of math ("raise both sides by e") here but simply definition.

 

[Symphonicity]

right, ok. That kind of makes sense, I guess it's just one of those processes I have to remember. Thank you. There will be more!

I haven't had proper teaching on this subject unfortunately so it's shaky as hell and I have very limited time to learn it before an exam.

When I want to eliminate "ln" from one side, I have to make the other side a power of "e".

 

[cutemouse]

Lies.

Let x = a+ib (where a and b are real)

Then e^(a)e^(ib)=-1

e^a (cos b + i sin b) = -1

Equate real and imaginary parts:

cos b = - e^(-a)

and sin b = 0

So b = π

and -1 = -e^(-a)
e^a = 1
a = 0

Thus, x = iπ is also a solution.

 

[Carrotsticks]

Lies.

Let x = a+ib (where a and b are real)

Then e^(a)e^(ib)=-1

e^a (cos b + i sin b) = -1

Equate real and imaginary parts:

cos b = - e^(-a)

and sin b = 0

So b = π

and -1 = -e^(-a)
e^a = 1
a = 0

Thus, x = iπ is also a solution.

I am very well aware of the complex solutions of the exponential function =)

If OP is struggling with solving exponential functions within R, do you really think it is appropriate to tell them of the solutions in C \ R ?

 

[cutemouse]

If OP is struggling with solving exponential functions within R, do you really think it is appropriate to tell them of the solutions in C \ R ?

A lie is still a lie.

Perhaps "no real solution" would've been more appropriate.

 

[seanieg89]

A lie is still a lie.

Perhaps "no real solution" would've been more appropriate.

The domain of interest is obvious from the forum the question was posted in.

If I asked you to solve some exponential equation without stating which field/ring to quantify over, and then called you a liar for not including quaternionic solutions, that would be equally ridiculous.

Of course a properly written question of this sort SHOULD specify domain, but at least in 2/3U there is absolutely no ambiguity whatsoever...we are always working in R.

 

[Sy123]

What does that mean? I see it a lot, but I fail to see the algebra manipulation. I learnt to read/comprehend "ln a = b" as "e to power b equals a" - there is no use of math ("raise both sides by e") here but simply definition.

I don't know if that is the official jargon for doing:



to raise both sides by 10 is



But I use those words to explain it

right, ok. That kind of makes sense, I guess it's just one of those processes I have to remember. Thank you. There will be more!

I haven't had proper teaching on this subject unfortunately so it's shaky as hell and I have very limited time to learn it before an exam.

When I want to eliminate "ln" from one side, I have to make the other side a power of "e".

Just think about what a logarithm means, a logarithm is the opposite of exponentiation, so to get rid of a logarithm, we simply exponentiate, this is the same logic when doing:



What do we want to get rid of? We want to get rid of a multiplication of x, and to 'undo' this process, we simply 'divide' by x.

The point I'm trying to make here is, don't simply take it as a rote learnt concept, understand why you are doing what you are doing.

 

[seanieg89]

I don't know if that is the official jargon for doing:



to raise both sides by 10 is



But I use those words to explain it

I would probably use the words: "raise 10 to the power of both sides" rather than "raise both sides by 10", but these are of course just words. The idea is much more important.

Would rep for the latter paragraph on the idea of "inversion" but have not spread it around enough.

 

[Symphonicity]

Hi guys,

I have another one - 3e^3t = 9 ?
I end up with 3t = ln3. Is that correct? And t = ln3/3?

 

[HeroicPandas]

Hi guys,

I have another one - 3e^3t = 9 ?
I end up with 3t = ln3. Is that correct? And t = ln3/3?

yes