Quick help with natural logs

e^(2X) = y-3

so if we take natural logs of both side we get:

ln[e^(2X)] = ln(y-3)

but i don't understand why the logs aren't taken like this:

ln[e^(2X)] = ln(y)-ln(3)

Could someone please explain?

Thanks :smile:

Reply 1

C.H.

1

Because e^2x is equal to y-2, the log of e^2x is the same as the log of y-2.

ln y - ln 2 is the same as ln (y/2), which is different altogether.

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Reply 2

C.H.

1

(accidentaly used 2 instead of 3 but same thing applies).

This was posted from The Student Room's iPhone/iPad App

Reply 3

raheem94

13

Original post by Nodes Of Ranvier

e^(2X) = y-3

so if we take natural logs of both side we get:

ln[e^(2X)] = ln(y-3)

but i don't understand why the logs aren't taken like this:

ln[e^(2X)] = ln(y)-ln(3)

Could someone please explain?

Thanks :smile:

We need to take the log of both sides not term by term.

E.g. if you have

A+B+C = D+E+F

When we take the logs of both sides we get

ln(A+B+C) = ln(D+E+F)

Reply 4

hellie7

2

because ln(y)-ln(3) is the same as ln(y/3) which is something completely different! the y-3 is what you want to take logs of! imagine it in brackets

Reply 5

Nodes Of Ranvier

OP

5

Original post by C.H.

Because e^2x is equal to y-2, the log of e^2x is the same as the log of y-2.

ln y - ln 2 is the same as ln (y/2), which is different altogether.

This was posted from The Student Room's iPhone/iPad App

I think I'm overcomplicating things, I see why now, thanks.

(edited 11 years ago)

Quick help with natural logs

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