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a level math logs question

It is given that ln(y+5)-lny=2lnx. Express y in terms of x, in a form not involving logarithms.

This is what I have so far (I can't upload a pic)
ln(y+5)-lny=lnx^2
lny+ln5-lny=lnx^2
Reply 1
Using log rules: subtraction of ln(y+5)-lny is the same as ln((y+5)/y). Then do everything to the power of e to get rid of ln. Then you should be able to rearrange for y.
Original post by zolearns
It is given that ln(y+5)-lny=2lnx. Express y in terms of x, in a form not involving logarithms.

This is what I have so far (I can't upload a pic)
ln(y+5)-lny=lnx^2
lny+ln5-lny=lnx^2

your expansion for ln(y+5) into lny+ln5 is incorrect. "lny+ln5" is ln(5y).

in this case leave the ln(y+5) as it is. logarithm law states when you subtract lns, you divide things in the function so you're supposed to do (y+5)/y. when you take e from both sides of the equations, the ln cancels out on either side and you just do simple rearranging.
Reply 3
Original post by h4nn4
your expansion for ln(y+5) into lny+ln5 is incorrect. "lny+ln5" is ln(5y).

in this case leave the ln(y+5) as it is. logarithm law states when you subtract lns, you divide things in the function so you're supposed to do (y+5)/y. when you take e from both sides of the equations, the ln cancels out on either side and you just do simple rearranging.


Be careful with your language here. You don't "take e from both sides of the equation" - you either "exponentiate both sides", or equivalently, "raise e to the power of each side" :smile: In effect, you're applying the inverse function to ln.
Original post by davros
Be careful with your language here. You don't "take e from both sides of the equation" - you either "exponentiate both sides", or equivalently, "raise e to the power of each side" :smile: In effect, you're applying the inverse function to ln.


very true, thanks

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