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solve for y: ln(y-2)-lnx+x =ln6
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add, base it off, and add again ....
\[\ln (y-2)-lnx+x=\ln6\] \[\ln \frac{ (y-2) }{x }=\ln6-x\] Taking the left and right hand side to the power of e, we get,
\[\frac{ (y-2) }{ x }=e ^{\ln6-x}\] \[\frac{ y-2 }{ x }=\frac{ 6 }{ e ^{x} }\] \[y=\frac{ 6 }{ e ^{x} }+2\]
let me know if i did something wrong
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