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# Core 3 - Logs

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1. if ln( x^6 y) - 2 ln (xy) + ln (3xy^2) = 2

prove that y = kx^n where k and n are constants

find the exact values of y = kx ^n

1) i've separated the equation:

ln x^6 + lny - 2(lnx + lny) + ln 3x + lny^2 =2

where do I go from here?
2. (Original post by Custardcream000)
if ln( x^6 y) - 2 ln (xy) + ln (3xy^2) = 2

prove that y = kx^n where k and n are constants

find the exact values of y = kx ^n

1) i've separated the equation:

ln x^6 + lny - 2(lnx + lny) + ln 3x + lny^2 =2

where do I go from here?
No need to separate them. Get them all under the same logarithm instead.
3. (Original post by Custardcream000)
if ln( x^6 y) - 2 ln (xy) + ln (3xy^2) = 2

prove that y = kx^n where k and n are constants

find the exact values of y = kx ^n

1) i've separated the equation:

ln x^6 + lny - 2(lnx + lny) + ln 3x + lny^2 =2

where do I go from here?
I think you've headed in the wrong direction to start with... kind of. Doesn't mean you can't go on from there, but you've kind of made... 1 more step... I dunno. Maybe that doesn't matter

Think about somehow combining the logs.
4. (Original post by RDKGames)
No need to separate them. Get them all under the same logarithm instead.
so do you get ln( x^6y 3xy^2 / x^2 y^2 ) =2

ln(3x^5y)=2

where do i go from here??
5. (Original post by Custardcream000)
so do you get ln( x^6y 3xy^2 / x^2 y^2 ) =2

ln(3x^5y)=2

where do i go from here??
ln is log with base e. So anti-log both sides.
6. (Original post by RDKGames)
ln is log with base e. So anti-log both sides.
3x^5 y = e^ 2

y= e^2 / 3x^5
7. (Original post by Custardcream000)
3x^5 y = e^ 2

y= e^2 / 3x^5
Yep. You can take the x out of the denominator as
8. (Original post by RDKGames)
Yep. You can take the x out of the denominator as
so y = (e^2/3) x^-5
9. (Original post by Custardcream000)
so y = (e^2/3) x^-5
Yep.
10. (Original post by RDKGames)
Yep.

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