Question

Algebra 2 please help!
Part 1: Is the following equation true? (2 points)

Part 2: Use complete sentences to explain the properties used in making your decision. (6 points)

4 log 3 x + one–half log 3 y – 2 log 3 z = log 3 x to the fourth power times square root of y all over z–squared.

Answer 1

The rules are as follows: \[\log _{a}(x \times y)=\log _{a}(x) + \log _{a}(y)\]AND\[\log _{a}(x)-\log _{a}(y)=\log _{a}\frac{ x }{ y }\]So let's see what we can do with that.

Answer 2

I really don't understand how to apply this I've been looking at the rules but I don't udneratand

Answer 3

\[\Large\rm 4\log_3 x + \frac{1}{2}\log_3 y – 2\log_3 z = \log_3 \left(\frac{x^4\sqrt y}{z^2}\right)\]This is what we're trying to determine, yes?
Did I type that correctly?

Answer 4

Ok, there's another one that deals with exponents, too. Then we will work on solving it.\[\log _{a}(x)^{p}=p \times \log _{a}(x)\]So let's go through this one thing at a time, if we can.

Answer 5

yes @zepdrix

Answer 6

The 4 is out front because, according to our rules, it was x's exponent. So the first part is \[\log _{3}(x)^{4}\]

Answer 7

The 1/2 in front of the second term was at one time\[\log _{3}(y)^{\frac{ 1 }{ 2 }}\]which is also \[\log _{3}(\sqrt{y})\]

Answer 8

So far we have\[\log _{3}x ^{4}+\log _{3}\sqrt{y}\]The addition sign in between them tells us that to simplify they are to be multiplied so\[\log _{3}(x \times \sqrt{y})\]

Answer 9

Oops that's

Answer 10

\[\log _{3}(x ^{4}\times \sqrt{y})\]

Answer 11

Now to deal with the 2 log base 3 of z...the 2 was once an exponent, according to our rules. So put it back there such that\[\log _{3}z ^{2}\]

Answer 12

Now we are ready to finish it off. The subtraction sign in between tells us that this is the denominator of a fraction. So the whole thing is, according to our rules of logs with the same base, is\[\frac{ \log _{3}(x ^{4}\times \sqrt{y}) }{ \log _{3}z ^{2} }\]

Answer 13

what about the other two terms?

Answer 14

do you do nothing with them?

Answer 15

I mean the last one sorry

Answer 16

@IMStuck @zepdrix I really need hep finishing this please help

Answer 17

whut? :U

Answer 18

can you please help finish off where he left ?

Answer 19

I don't see where you left off..
around here?\[\Large\rm \log_3\left(x^4 \sqrt{y}\right) – 2\log_3 z = \log_3 \left(\frac{x^4\sqrt y}{z^2}\right)\]

Answer 20

\[\Large\rm \log_3\left(x^4 \sqrt{y}\right) – \log_3 z^2 = \log_3 \left(\frac{x^4\sqrt y}{z^2}\right)\]

Answer 21

no he last equation he put in
log v3 (x^4 x sqrt y)/log v3 z^2

Answer 22

\[\Large\rm \log_3\left(\frac{x^4 \sqrt{y}}{z^2}\right)= \log_3 \left(\frac{x^4\sqrt y}{z^2}\right)\]Here....?
Then the problem is finished.
See how you've shown that both sides are the same?

Answer 23

how did you get that?

Answer 24

I don't know why he wrote log_3 z^2 in the bottom, that must've been a typo.

Answer 25

This is our log rule:\[\Large\rm \color{royalblue}{\log(a)-\log(b)=\log\left(\frac{a}{b}\right)}\]Not this:\[\Large\rm \color{red}{\log(a)-\log(b)\ne\frac{\log(a)}{\log(b)}}\]

Answer 26

Maybe color will help with the rule, hold on one sec :d

Answer 27

\[\Large\rm \log_3\left(\color{orangered}{a}\right)-\log_3\left(\color{royalblue}{b}\right)=\log_3\left(\frac{\color{orangered}{a}}{\color{royalblue}{b}}\right)\]Applying this to our problem:\[\Large\rm \log_3\left(\color{orangered}{x^4 \sqrt y}\right)-\log_3\left(\color{royalblue}{z^2}\right)=\log_3\left(\frac{\color{orangered}{x^4 \sqrt y}}{\color{royalblue}{z^2}}\right)\]Understand how the rule gets applied?

Answer 28

why is a x4 sqrt 7 and why is b z2

Answer 29

Because the subtraction is in front of the b term.

Answer 30

The one that has the subtraction is the one that ends up in the denominator.

Answer 31

okay i kind of get it lol thank you!

Answer 32

c:
math is fun! don't give up!! lol

Answer 33

a bit late lol

Answer 34

haha XD

Answer 35

I m so sorry......my computer totally crashed! Im glad you got the help you needed to finish, though. Did not mean to leave you hanging there!