# Integrate ln

1. Dec 9, 2009

### ibysaiyan

1. The problem statement, all variables and given/known data
Find area of region bounded by curve with equation y=e^2x , the x-axis and the lines x=-ln3 and x=-ln2.

2. Relevant equations
log. law + integration

3. The attempt at a solution
well here is how i started this:
y=e^2x after integration
(1/2e^2x)
=>(1/2e^-(ln6)-(1/2^-(ln4) )
=>(1/2e^1/6) - (1/2e^1/4)
=>(1/2e^1/6)/ (1/2e^1/4)
=>(1/6(lne)1/2) / (1/4(lne)1/2)
(1/6X1/2) / (1/4X1/2)
=> is this correct?

2. Dec 9, 2009

### ideasrule

e^(-ln6) is not equal to e^1/6; it's equal to e^ln(1/6). What happens to any number if you take its natural log and raise e to the result?

3. Dec 9, 2009

### ibysaiyan

y=lnx => x=e^y ?

Last edited: Dec 9, 2009
4. Dec 9, 2009

### ibysaiyan

still confused... but i thought just as -ln3 = 3^-1 = 1/3 ,same would happen to this but guess thats not the case ?

5. Dec 9, 2009

### ideasrule

It is the case, but you wrote:

1/2e^-(ln6)

then, in the next step,

(1/2e^1/6)

That would mean -(ln6) is equal to 1/6, which it isn't. Instead, -(ln6) is equal to ln(1/6).

6. Dec 9, 2009

### ibysaiyan

ah k, now i get it yea.. 1/6 would only be equal to 6^-1 , erm.. sorry for being such a nuisance but now what to do with the ln(1/6)? , i cant ln both sides .. =/