A level logarithms help

the question is lnx+ln2x = 12

I used a log law and got ln2x^2 which I then wrote as 2ln2x = 12
2loge2x=12
loge2x=6
e^6=2x
this gives me x=201.7
The answer is 285. Where did I go wrong?

Reply 1

davros

Original post by epic08

the question is lnx+ln2x = 12

I used a log law and got ln2x^2 which I then wrote as 2ln2x = 12
2loge2x=12
loge2x=6
e^6=2x
this gives me x=201.7
The answer is 285. Where did I go wrong?

How did you convert from the 1st bold expression to the 2nd one?

I would write ln(2x) = ln2 + lnx so your original equation becomes 2lnx + ln 2 = 12 and proceed from there :smile:

Reply 2

ghostwalker

Original post by epic08

the question is lnx+ln2x = 12

I used a log law and got ln2x^2 which I then wrote as 2ln2x = 12
2loge2x=12
loge2x=6
e^6=2x
this gives me x=201.7
The answer is 285. Where did I go wrong?

ln 2x^2 - yep, combining the logs, good move.

=2ln2x - no. You're thinking, I'd guess, ln y^2 = 2 ln y. But here you have 2 times x^2, not (2x)^2, so you can't pull out the exponent.

From ln 2x^2=12, I'd then take the exponentials at that point.

Edit: This is a different route to the one davros is suggesting, so choose one - either will do.

(edited 2 years ago)

Reply 3

epic08

OK YES I have got the right answer now

ln2+2lnx = 12
2lnx=11.3 (subtracted ln2)
lnx=5.65
logex=5.65
so x=e^5.65
x=285

Thank you so much I understand where I went wrong now!!!

Reply 4

ghostwalker

Original post by epic08

OK YES I have got the right answer now

ln2+2lnx = 12
2lnx=11.3 (subtracted ln2)
lnx=5.65
logex=5.65
so x=e^5.65
x=285

Thank you so much I understand where I went wrong now!!!

The route @davros suggested is the better of the two, as it avoids x^2, which would require a little extra thought when taking the square root at the end.