Solve ln y - ln (y - 4) = ln 2?

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Solve ln y - ln (y - 4) = ln 2?

Postby Integr88 » Sat 02 April, 2011 18:05

solve ln y - ln (y - 4) = ln 2 for y
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Re: Solve ln y - ln (y - 4) = ln 2?

Postby Integr88 » Sat 02 April, 2011 18:14

we have ln y - ln (y - 4) = ln 2
using ln rule ln(a)-ln(b)=`ln(a/b)` we can write:
`ln(y/(y-4))=ln2`

Now we using e we get rid of the ln:
`e^ln(y/(y-4)) = e^(ln 2)` the `e^ln(a)`=a. Using this rule we can write:

`y/(y-4)=2`
Now using cross-multiply we have:
`y=2(y-4)`
Now expand and we have `y=2y-8`

collect like terms with y on one side. Subtract y from both side and we have:
`y-y=2y-y-8`
`0=2y-y-8` now add 2y-y=y

`0=y-8` add 8 to both side `0+8=y-8+8` and we have:
`8=y`

so we found y to be 8
`:.` y=8
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