Question

Solve for y. ln(y-1)-ln2=x+ln x

Answer 1

ln(y-1)-ln(2)-lnx=x

use this:
ln(a)-ln(b)-ln(c)=ln(a)-[ln(b)+ln(c)]=ln(a)-ln(bc)=ln[a/(bc)]
and write x as lne^x

Answer 2

so then you would have
\[\frac{a}{bc}=e^x\]

Answer 3

Uhmm..
e^(ln(y-1/2)) = e^(x+ln x) ??

Answer 4

no i wrote everything with a ln on one side did you what i wrote above?

Answer 5

\[y=1+e^{x+\ln(2x)}\]

Answer 6

\[\frac{y-1}{2x}=e^x\]

Answer 7

\[y-1=2xe^x\]
\[y=2xe^x+1\]

Answer 8

Oh okay I see now! So log and ln have the same properties?

Answer 9

yes
they are both log
\[\ln(x)=\log_e(x)\]

Answer 10

that subscript is tiny but its suppose to be an e

Answer 11

Yes that is an e

Answer 12

ok i couldn't read my own typing