OpenStudy (anonymous):

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

6 years ago
hartnn (hartnn):

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

6 years ago
OpenStudy (anonymous):

Yes I do

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

How do I plug that in

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

Log(z^3)?

6 years ago
OpenStudy (anonymous):

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

6 years ago
hartnn (hartnn):

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

6 years ago
OpenStudy (anonymous):

ok... thanks.... :)

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

That's quite alright! :)

6 years ago
OpenStudy (anonymous):

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?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

here are the properties: \(\large log_bx + log_by = log_b(x\cdot y) \) and \(\large log_bx - log_by = log_b(\frac{x}{y}) \)

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

Ohh ok

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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?

6 years ago
OpenStudy (anonymous):

Yess

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

So that's the simplifed expression o.O

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

what did you get?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

|dw:1345235816664:dw|

6 years ago