Question

Write in simplified form 2log x+log y-3logz

Answer 1

do you know properties of logarithm??
like this log a + log b = log (a*b)

Answer 2

Yes I do

Answer 3

use this property first: \(\large log_b(x^y)=ylog_b(x) \)

Answer 4

How do I plug that in

Answer 5

then use the product to sum and quotient to difference properties.

Answer 6

in your first term,
\(\large 2logx=log(x^2) \)
can you do that with the third term in your equation?

Answer 7

Log(z^3)?

Answer 8

yes...
@hartnn , sorry, i didn't mean to hijack the problem.... do you want to continue this?

Answer 9

no u didn't hijack.....u please continue :)

Answer 10

ok... thanks.... :)

Answer 11

sorry haley.... chrome crashed on me....:(

Answer 12

That's quite alright! :)

Answer 13

so now your expression is: \(\large log(x^2)+logy-log(z^3) \)

now can you do the sum to product and difference to quotient rules for logs from here?

Answer 14

It's new material so I'm not very familiar with it. How do I begin?

Answer 15

here are the properties:

\(\large log_bx + log_by = log_b(x\cdot y) \) and
\(\large log_bx - log_by = log_b(\frac{x}{y}) \)

Answer 16

Ugh so lost I'm still not quite sure how to plug those in to solve

Answer 17

ok... i shouldn't have used x's and y's in the properties because your equation uses the same variables.....
lemme try again....

Answer 18

Ohh ok

Answer 19

What did you get for your answer? I got logx+logz
never mind.. that can't be right

Answer 20

this is your expression:
\(\large log(x^2) +logy - log(z^3) \)
applying the sum to product property on the first two terms we get:

\(\large \color {red}{log(x^2) +logy} - log(z^3) \)
\(\large \color {red}{log(x^2\cdot y)} - log(z^3) \)

still with me?

Answer 21

Yess

Answer 22

now applying the difference to quotient property of logs on the remaining terms to simplify we get:
\(\large log(x^2\cdot y) - log(z^3) \)
\(\large log(\frac{x^2\cdot y}{z^3} ) \)

Answer 23

So that's the simplifed expression o.O

Answer 24

I got something completely different lol. But you really helped out.

Answer 25

yep... your original expression was expressed in terms of 3 logs this one has 1...

Answer 26

what did you get?

Answer 27

i think i'm correct... lemme check...