Logarithms/Natural Log's: Finding the Inverse Equation

[Guest]

I've been getting through most of these problems pretty well, but I'm stuck on two that I just can't seem to figure out.

Here they are:

y=(1+e^(x))/(1-e^(x))

and

f(x)=2-e^(x)

I need to find the inverse equations... But the answers I come up with aren't correct.

Could someone please show me how to work these two?

Thanks.

 

[Unco]

Show your efforts.

I take it you know to swap x and y around and solve for y.

After doing so for the first one, multiply both sides by the denominator and group e^y terms together to solve for e^y.

Remember for both that taking natural logs undoes e^y.

 

[Guest]

That's what I had attempted to do with both of them... Maybe I'm just forgetting some rules?

I get x-e^(y)x=1+e^(y) for the first one. Then I can subtract 1 from the right side and take the natural log of e^(y) to get y, but then I stop because I'm pretty sure you're not allowed to take ln(x-e^(y)x) on the left side because there's more than one term? What do I do from x-e^(y)x=1+e^(y)?

I run into the same problem with the second one, I rearrange it to get e^(y)=2-x and then stop because there is more than one term on the right side of the equation.

Also, sorry to expand on this post, but I have a few more problems concerning logarithms/natural logs:

I have 2^(log23+log25). I'm attempting to find the exact value here. So I rearrange and get log2(15)ln(2). Where do I go from there?

And the last one:

e^(ax)=Ce^(bx), where a is not equal to b. Here I'm solving for x, and I don't even know how to start it.

I really, REALLY appreciate any help I can get from here! Thanks.

 

[Unco]

I get x-e^(y)x=1+e^(y) for the first one. Then I can subtract 1 from the right side and take the natural log of e^(y) to get y, but then I stop because I'm pretty sure you're not allowed to take ln(x-e^(y)x) on the left side because there's more than one term? What do I do from x-e^(y)x=1+e^(y)?

Group the e^y terms to one side:

xe^y + e^y = x - 1

And factor out e^y:

e^y(x + 1) = x - 1

Now you can solve for e^y, and hence y.

I run into the same problem with the second one, I rearrange it to get e^(y)=2-x and then stop because there is more than one term on the right side of the equation.

Just as e^y = x --> y = ln(x),

e^y = 2 - x --> y = ln(2 - x)

We have taken the natural log of both sides of the equation; it doesn't matter how many terms there are.

I have 2^(log23+log25). I'm attempting to find the exact value here. So I rearrange and get log2(15)ln(2). Where do I go from there?

That looks a little dodgy.

Don't forget the fundamentals:

\(\displaystyle \mbox{ x^{m+n} = x^m \times x^n}\)

(Alternatively -- well, not really -- you could consider the log of a product rule to simplify the exponent.)

and \(\displaystyle \mbox{b^{(\log_b{x})} = ?}\)

e^(ax)=Ce^(bx), where a is not equal to b. Here I'm solving for x, and I don't even know how to start it.

This is a good finaliser.

Does taking natural logs of both sides sound familiar?

Next I would like you to consider everything that has been covered:

What rules can we use with ln(Ce^(bx))?

And what do we do to solve for x when we have an ax and a bx term? (similar process to how we solved for e^y in your first question)

 

[Gene]

#1 You are taking the ln too soon.
x=(1+e^y)/(1-e^y)
x-x*e^y=1+e^y
x-1=e^y+x*e^y=e^y(x+1)
e^y=(x-1)/(x+1)
Now ln gives
y=ln((x-1)/(x+1))


#2 I see nothing wrong with
y=ln(2-x)

#3 Not sure what you mean by exact value. Mixing log & ln doesn't seem kosher and I don't follow your rearangement.
2^(log(23*25))

#4 A good start would be
ax=ln(C)+bx

P.S. Sorry Unco. You weren't here when I started.