Question

Need help to write the expression as a single logarithm with a coefficient of 1

Answer 1

\[\log_{10} (x ^{2}-16)-3\log_{10}(x+4)+2\log_{10}x \]

Answer 2

i need to understand how to do this.. not just the answer.

Answer 3

You need two properties of logs
\[\log_{b}M + \log_{b}N=\log_{b}MN\]
and
\[p \log_{b}M=\log_{b}M^{p}\]

Answer 4

Oh yeah, following from the first property
\[\log_{b}M - \log_{b}N=\log_{b}\frac{M}{N}\]
provided N is not zero

Answer 5

Actually, provided N > 0

Answer 6

\[\log_{10}(x ^{2} -16)-\log_{10}(x+3)^{3} +\log_{10}x^2 \]

Answer 7

So far so good. Now combine them all into one log. Hint: If it's a positive log, it goes in the numerator, if it's negative, it goes in the denominator...

Answer 8

\[\frac{ \log_{10}(x ^{2}-16) }{\log_{10}(x+4)^{3}\times x ^{2} }\]
I get confused with the cube and the square. and i probably need to expand the numerator into (x+4)(x-4)??

Answer 9

Wait, let's go over this again... :)

Answer 10

\[=\log_{10}\frac{ 1 }{(x+4+x ^{2}) } \]?

Answer 11

Once you combine them into one expression, only one log is needed...
also, if the log was positive, then it goes in the numerator.

Answer 12

so x^2 goes in the numerator as well?

Answer 13

Yes it does.

Answer 14

And only one log

Answer 15

do you expand out the numerator and denominator?

Answer 16

Step by Step:
\[\log_{10}(x ^{2} -16)-\log_{10}(x+3)^{3} +\log_{10}x^2\]
\[\log_{10}\frac{x^{2}(x^{2}-16)}{(x+3)^{3}}\]
this part, you get, right?

Answer 17

And if you don't, don't hesitate to tell me...

Answer 18

yes i do

Answer 19

Well, too bad, I made a serious typo (and so did you) :P
I'm sorry, it should be
\[\log_{10}(x ^{2} -16)-\log_{10}(x+4)^{3} +\log_{10}x^2\]
\[\log_{10}\frac{x^{2}(x^{2}-16)}{(x+4)^{3}}\]

Answer 20

So, there, from there, you can do your fancy factoring and cancellations :)

Answer 21

What's important is that it's now a single log with coefficient 1,as was required.

Answer 22

ahh i change it to x+3 heh. ok.

\[\log_{10}\frac{ (x-4)x^{2} }{ (x+4)^{2} } \]

hmm this isnt right..

Answer 23

Why not?

Answer 24

it doesnt seem simplified enough... ?

Answer 25

It is.

Answer 26

Unless you prefer to distribute the x^2, and evaluate the square of the binomial at the bottom.
Personally, I like to leave it in factored form.

Answer 27

so that is the final answer???!!

Answer 28

so if its \[\log_{10}(x ^{2} -4)^{2}-3[\log_{10}(x+4)+2\log_{10}x] \]
then i just do whats in the brackts first. so (x+4)(x^2) ?

Answer 29

yes, IF there are brackets.

Answer 30

ok, and thanks for your help, much appreciated.

Answer 31

No problem :D