Exponential & Logarithm Question Help - Please?

[rufusbabe]

Hi,

I have come across a question that has stumped me a little... here it is...

Express y in terms of x (I'm guessing that means rearrange to make y the subject?!):

x = 5e^y + 2e^y - e^(2+y)

This is what I've got so far (no idea if it's correct):
x = e^(y+5) + e^(y+2) - e^(2+y)

x = e^(y+5)

ln x = ln e^(y+5)

ln x = (y+5) ln e (ln e = 1 right?)

ln x = (y+5) 1

ln x = y + 5

ln x - 5 = y

Is this correct?? I'm really new to exponential and logs so any help/advice would be much appreciated.

Thank you!

 

[wjm11]

x = 5e^y + 2e^y - e^(2+y)

This is what I've got so far (no idea if it's correct):
x = e^(y+5) + e^(y+2) - e(2+y)

Your work seems to suggest that 5e^{y =} e^(y+5)
Hopefully, you have some misgivings about the truth of this. Let's test it; let y = 0 for simplicity: 5e^{0 =} e⁽⁰⁺⁵⁾ ==> 5 ⁼ e⁵
Obviously, this is not true.

Let's start anew:

x = 5e^y + 2e^y - e^(2+y)
x = 7e^y - e^(2+y)

x = 7e^y - e² * e^y

x = e^y (7 - e² )

Can you continue?

 

[rufusbabe]

Bummer... I knew when you had a number out the front of a log e.g. 2log3 it actually means squared or cubed or whatever i.e. log3² - I was hoping that would be the same with e but apparently not.

I'll have a go at continuing...

x = e^y (7 - e² )

x = e^y (7 - 7.389056099)

x = e^y (-0.389056099)

ln x = ln e^y * -0.389056099

ln x = y ln e * -0.389056099

ln x = y * 1 * -0.389056099

ln x = y * -0.389056099

ln x / -0.389056099 = y

Not confident....


Thank you so much for the help though! I really appreciate it.

 

[wjm11]

x = e^y (7 - e² )

x = e^y (7 - 7.389056099)

x = e^y (-0.389056099)

You were okay up to here. But in the next step, there is a problem. You must realize that when you take the ln of both sides, it is of the entire side, not just select items.

ln x = ln [e^y * -0.389056099]

We could proceed from here, but I'd recommend going back to x = e^y (7 - e² ) and isolating the term e^y

e^y = x/(7 - e² )

Now take the ln of both sides:

ln e^y = ln [x/(7 - e² )]

y = ln [x/(7 - e² )]

 

[rufusbabe]

Any chance you could explain how you went from

x = e^y (7 - e²)

to...

e^y = x/(7 - e²)

How does the x end up as the numerator? Sorry, my equation rearranging is a little rusty.

Thank you though!! I find all this maths so hard to get my head around.

 

[rufusbabe]

Ah ha! I see said the blind man!

Thank you.