Question

Use properties of logarithms to condense the logarithmic expression. Write the expression as a single logarithm whose coefficient is 1. Where possible, evaluate logarithmic expressions without using a calculator.

log(x)+log(x^2-9)-log15-log(x+3)

Answer 1

Hmm this one shouldn't be too bad.
It's pretty much what we did in the last problem, just in reverse.

Answer 2

yay :)

Answer 3

\[\Large\rm \color{orangered}{\log(x)+\log(x^2-9)}-\log15-\log(x+3)\]We'll condense these first two terms down using a rule of logs:\[\Large\rm \log(a)+\log(b)=\log(ab)\]

Answer 4

log(x(x^2-9))

Answer 5

When you bring the x into the log, make sure it multiplies the `entire` x^2-9, not just the x^2.
So adding another set of brackets might help.\[\Large\rm \color{orangered}{\log\left[x(x^2-9)\right]}-\log15-\log(x+3)\]Oh you're too quick for me c: very nice!

Answer 6

Then we'll apply our other rule of logs,\[\Large\rm \log(a)-\log(b)=\log\left(\frac{a}{b}\right)\] to these two terms:\[\Large\rm \color{#CC0033}{\log\left[x(x^2-9)\right]-\log15}-\log(x+3)\]

Answer 7

help :(( hahahahaha

Answer 8

I'm sorry

Answer 9

you -_-

Answer 10

So the rule is telling us we can condense the two logs down to a single log, and divide by the 15, yes?\[\Large\rm \color{#CC0033}{\log\left[\frac{x(x^2-9)}{15}\right]}-\log(x+3)\]

Answer 11

is it log(x(X^2-9)-log15/log(x+3)

Answer 12

hmm no :o
try to follow along missy!

Answer 13

sorry that sent late!

Answer 14

but okay im following you again continue :)

Answer 15

Oh laggy? :o that's no fun.

Answer 16

So we'll apply that same division rule to the last two terms.

Answer 17

We'll combine them into one log,
throwing the second term into the denominator.

Answer 18

\[\Large\rm \log\left[\frac{x(x^2-9)}{15(x+3)}\right]\]Something like that, yah? :d

Answer 19

yes yes !

Answer 20

is that the final answer?

Answer 21

So we've successfully smashed it down into a single log using our fancy rules.
And it doesn't have any coefficient showing in front, which means the coefficient is 1.
So yay! Looks like we did everything they asked for.

Answer 22

can you type that answer into this engine? I don't think I'm entering it how I'm suppose to

Answer 23

\[\Large\rm \log\left[\frac{x(x^2-9)}{15(x+3)}\right]\]
=log((x(x^2-9))/(15(x+3)))

Answer 24

its asking if anything inside the log can be factored or cancelled

Answer 25

oh oh oh. my bad.

Answer 26

We have the `difference of squares` in the numerator.\[\Large\rm x^2-9=x^2-3^2\]Difference of squares can be factored into the product of conjugates:\[\Large\rm a^2-b^2=(a-b)(a+b)\]

Answer 27

\[\Large\rm x^2-3^2=(x-3)(x+3)\]

Answer 28

And then it looks like we can cancel out our (x+3)'s, yes?

Answer 29

yes

Answer 30

\[\Large\rm \log\left[\frac{x(x-3)\cancel{(x+3)}}{15\cancel{(x+3)}}\right]\]

Answer 31

= log((x(x-3))/15)
Hopefully that works :o

Answer 32

so is the final answer than log(x(x-3)/15) ?

Answer 33

Yah that looks better.
I guess I put more brackets than was needed.

Answer 34

well thank you so so much :)

Answer 35

youre the best!