Re: Do I need logarithms to solve this?

ljs15 wrote: 
Hi all.

Firstly, apologies if this newsgroup isn't the place for asking questions
such as this. If that's the case just let me know.


Not really, but you've done it now. You'd be better with sci.math
really.

Right - I have the equation y = 2e^x+1, and I need to find x in terms of y.
Am I heading along the right lines by taking the natural logarithm of each
side, i.e:

ln y = ln(2e^x) ln(2e^1) 

I don't know why you think that ln(2e^1+1) is equal to that,
but it certainly isn't. Remember, ln(xy)=ln(x)+ln(y)---there
ins't a nice expresion for ln(x+y) in terms of ln(x) and ln(y).

However, if you get 2e^x on one side of the equation
and everything else on the other, you will have something
you can usefully take the logarithm of. Try that.
.

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