exponential

cazwhatever

Hi all

I want to find the exponential of

ln(10+x)+ln(4x+4) - 3x

I know that ln(10+x)+ln(4x+4) = ln (10+x)/(4x+4) and the exponential of that is (10x+x)/(4x+4)

I'm happy enough with that part but I'm unsure what to do with the -3x.
Is it
(10x+x)/(4x+4) +exp(-3x) or

(10x+x)/(4x+4) -exp(3x) or

(10x+x)/(4x+4) -exp(-3x)?

(by exp I mean "e to the power of...")

If you can even direct me to a website that could give me a clue that would be brilliant

tacofries

i couldnt tell u but if thats from exam papers this website should have the solution http://www.studentxpress.ie/papers.htm

**Timbuk2**

Here's two hints!

Firstly log(a) + log(b) = log(ab), you haven't that bit correct!
Secondly, when you get the exponent of the whole expression, [latex]e^{x+y} = e^xe^y[/latex] - think about how you can apply that to your expression.

Edit: In this case it might be more appropriate to think of [latex]e^{x-y} = \displaystyle\frac{e^x}{e^y}[/latex], but these are effectively equivalent!

cazwhatever

**Timbuk2** wrote: »

Here's two hints!

Firstly log(a) + log(b) = log(ab), you haven't that bit correct!
Secondly, when you get the exponent of the whole expression, [latex]e^{x+y} = e^xe^y[/latex] - think about how you can apply that to your expression.

Edit: In this case it might be more appropriate to think of [latex]e^{x-y} = \displaystyle\frac{e^x}{e^y}[/latex], but these are effectively equivalent!

Thank you sooo much!! Spent all day trying it out different ways and it was right in front of my eyes the whole time. Thank you so much