Integrating log functions (1 Viewer)

[Jacqui]

I am having big problems!
The question is...
If intergral (from 5 to 1) dx/2x-1 = ln z, find the value of z.

 

[Slice of heaven]

Originally posted by Jacqui
I am having big problems!
The question is...
If intergral (from 5 to 1) dx/2x-1 = ln z, find the value of z.

I think I may be able to help, here is my solution:

If integral (from 5 to 1)dx/2x-1, then just integrate as normal

=[1/2ln(2x-1) ](from 5 to 1)

=1/2ln(9)-1/2ln(1)

=1/2ln(9), as ln(1)=0

=ln(9to the pwer of half/ or square root of 9)using the dropping of the power rule in reverse, sorry bout that but im not good with computers(typing.. etc)

=ln(3)

Thus ln(3)=lnz, therefore z=3

 

[Lazarus]

Very nice.