Question

i need help badly!!!

Answer 1

\[2\log_{a} W-1/4 \log_{a}Z+7\log_{a}Y \]

Answer 2

I have to write it as a single Logarithm

Answer 3

http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif

Answer 4

rules 2, 1 and 3 listed there, would apply

Answer 5

what he said ^^^^

Answer 6

use rule 3 first on all 3 terms, then use rule 2 a few times

Answer 7

@zzr0ck3r im stuck now because i did the rules and now i have \[\log_{a}W ^{2}/Z ^{1/4}+\log_{a}Y ^{7} \]

Answer 8

@jdoe0001 ^^

Answer 9

Is this \[\frac{ \log _{a}W ^{2} }{ Z ^{1/4}+\log _{a}Y ^{7} }\] or \[\frac{ \log _{a}W ^{2} }{ Z ^{1/4} }+\log _{a}Y ^{7}?\]

Answer 10

the second one but z^1/4 is only being divided into w^2

Answer 11

@Dscdago

Answer 12

Wow...not to mess up the already messed up, but I got something here...\[\log _{a}\frac{ W ^{2}Y ^{7} }{ Z ^{\frac{ 1 }{ 4 }} }\]

Answer 13

i was just doing what people told me to do. @IMStuck

Answer 14

\(2\log_{a} W-1/4 \log_{a}Z+7\log_{a}Y\\\log_a(W^2)-\log_a(Z^{\frac{1}{4}})+\log_a(Y^7)\\\log_a(\frac{W^2}{Z^{\frac{1}{4}}})+\log_a(Y^7)\\\log_a(\frac{W^2}{z^{\frac{1}{4}}}Y^7)\)

Answer 15

I get it. These stink. They are so confusing. When you have one log subtracted from another, the one that is being subtracted goes under the first as a fraction. Hence the reason I put log base a of Z to the 1/4 under the log base a of W^2. Now the W is squared because in its original form it is W^2, then the 2 comes down in front like you have it in your unsimplified version. \[\log _{a}W ^{2}=2\log _{a}W\]See that?

Answer 16

all you need is the one more step using the fact that \(\log_a(x)+\log_a(y)=\log_a(xy)\)

Answer 17

zzrOck3r the y^7 is up on top of the fraction with the W^2.

Answer 18

it doesn't really matter if the Y is up top or on the side because it means the same thing.

Answer 19

correct

Answer 20

Yes it does mean the same thing but it doesn't look as organized in "normal" ways of simplfying these dumb things.

Answer 21

\(a*\frac{b}{c}=\frac{ab}{c}\)

Answer 22

so we can write the final answer as \(\log_a(\frac{W^2Y^7}{Z^{\frac{1}{4}}})\)

Answer 23

or even \(\log_a(W^2Y^7Z^{-\frac{1}{4}})\)