Question

Algebra 2 help please

Answer 1

2.G
3.C

1,4,5 are what i need help with.

Answer 2

ok lets do number 1 :)

Answer 3

\[clog_{b}a=\log_{b}a ^{c}\]
using this formula, can u tell me wat u think is the answer to number 1?

Answer 4

D

Answer 5

yes :)

Answer 6

can u do number 4 now?

Answer 7

G

Answer 8

no, now given this
\[4\log_{3}x+7\log_{3}y = \log_{3}x ^{4} + \log_{3}y^{7}\]
try again :)

Answer 9

H

Answer 10

the answer is F

Answer 11

beacause \[\log_{b}a+\log_{b}c=\log_{b}ac\]
understand? also when ur back plz stay so we can finish this lol

Answer 12

ohhh i get it @HawkCrimson

Answer 13

cool now number 5 :D

Answer 14

i can completely understand y this one is giving u trouble cause they phrase it so wierd lol so ill just summerize wat the entire thing means ;)

Answer 15

\[-\log10^{-7}-(-\log10^{-14})\]
Try to do number 5 part a now ;)

Answer 16

i keep getting nan

Answer 17

i know it says don't use a calculator,but i can't do it without it

Answer 18

lol that is because they want u to simplify it

Answer 19

ill show u how now

Answer 20

\[-\log10^{-7}+\log10^{-14}\]
\[\log10^{-14}-\log10^{-7}\]
\[\log(10^{-14}/10^{-7}) = \log10^{-7}\]

Answer 21

\[\log10^{-7}\] \[-7\log_{10}10\] Since, \[\log_{a}a=1\] Then, \[\log_{10}10 = 1\]
Therefore, -7(1) = -7
-7 is ur answer :)

Answer 22

understand?

Answer 23

ohhhhhh i get it

Answer 24

cool now part b, u give it a try :)

Answer 25

oh lol i cannt help with part b really, just explain wat i did in words and if u want i can check it :)