Some Logs/exponential help! (1 Viewer)

0hNivlek · Feb 4, 2009

Hey guys~ Just confused on some log/exponential stuff. How do i do these?

<o>thanks!</o>

Trebla · Feb 4, 2009

0hNivlek · Feb 4, 2009

!!!!! Thank you so much! Gah thanks!!!!!!!!

0hNivlek · Feb 4, 2009

Quick question. I kinda don't understand the part in question 25 where 1 + ln x = 0. Like how do you solve that? How and why does it become x = e ^ (-1)?

Thanks!

shaon0 · Feb 4, 2009

0hNivlek said:
Quick question. I kinda don't understand the part in question 25 where 1 + ln x = 0. Like how do you solve that? How and why does it become x = e ^ (-1)?

Thanks!

This is what Trebla did.
f'(x)=x(1/x)+1(ln(x))
=> x(1/x)=1
=> derivative of ln(x)=1/x, derivative of x=1.
Let f'(x)=0.
ln(x)+1=0
ln(x)=-1
Since, the exponential function is the inverse of the natural logarithm function you can use it to get 'rid' of the ln(x).
So Trebla took the exponent of both sides.
=> e^(ln(x))=e^(-1)
x=e^(-1).

0hNivlek · Feb 6, 2009

^ Thanks for the explanations! Greatly appreciated

!

Anyways, i got some more problems ;x. I did the whole excerice but got stuck on these. It's from the HSC Jones & Couchman 2U book btw page 247.

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Thanks!

lolokay · Feb 6, 2009

if ln_a(-4) = b
then a^b = -4
but a positive number to any real power gives you a number > 0, so unless you're going into complex numbers, the logarithm isn't meaningful

for 21. it's easiest to use e^lnx = x
so for a) x = ln2

shaon0 · Feb 6, 2009

lolokay said:
if ln_a(-4) = b
then a^b = -4
but a positive number to any real power gives you a number > 0, so unless you're going into complex numbers, the logarithm isn't meaningful

for 21. it's easiest to use e^lnx = x
so for a) x = ln2

a^b=-4
b.ln(a)=ln(-4)
b.ln(a)=ln(-1)+ln(4)
b.ln(a)=i.pi+ln(4) [since, e^(i.pi)=-1]

Some Logs/exponential help! (1 Viewer)

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