Help with Logarithm?

Hey guys can anyone give me some pointers on how to go about solving the following?

I tried a) but think I messed up big style?






For a) I started (after some guidance from an internet source)...


log(4x)=7.2

log(4x)=36/5

4x=e^36/5

x=e^36/5
........4



So, completely wrong?

So, completely wrong?

Are you expected to come out with a decimal answer to a certain number of places?

For (a) I would start by using the standard law of logs: log(ab) = log a + log b

Thanks, I'll go back and flick through my notes on all the laws of logarithms. Its getting too much for me.


Doing my degree in BioMedical Science, which Im good at. But they're making us do all this Maths, which is of no use to my degree specifically including the log stuff. We've got 8 weeks to learn it at 3 x50 minute lectures per week then a formal examination at the end.


Stressed! haha

No

But you could have just left it as

$x = \dfrac{e^{7.2}}{4}$

Thanks so I should write my answer as


a) log(4x)=7.2

4x=e^7.2

x=e^7.2
.......4





At first I wasn't sure if I had to work it out with Log10. Like this..

a) log(4x)=7.2


10^7.2=4x


x = 10^7.2
.........4


15848931.92
.......4

x=3962232.98

I'd actually imagine that logs would be quite useful for describing laws of biology and chemistry!! Keep at it - these questions aren't too bad when you've done a few.

sorry you are correct ... so tuned into ln

$x = \dfrac{10^{7.2}}{4}$

Im going to tackle question B and I shall be back in a second with an answer.

ok

b) in(2x)=5.2


x = e^5.2
.........2


e^5.2=181.2722419


181.2722419
........2

=90.63612094.......



x=90.63612094

That's good right? Haha.


Well happy now. I shall attempt C and D together and will be back! Muahaha.

usually if log is written without a base then it's base 10 - so is lg

ln, as you know is base e

so to solve log 4x = 7.2 you quite rightly say 4x = 10^7.2 etc

and for ln 2x = 5.2 2x = e^5.2 etc - your answers are right

good luck with parts c and d !!

Do I subtract the powers from one another?

c) in(x^3)-in(x^2)=1

yes


$ln(x^3)-ln(x^2) = ln(\frac{x^3}{x^2}) = lnx = 1$

so

x = e


NOTE

It is ln, not in