## swiftskier96 Help!?!?!?!? For questions 1 and 2, write the expression as a single logarithm. (Problem written below) logb (q^2 + t^8) logb(q^2t^8) logb(qt^(2+8)) In this answer, the 2+8 is the entire exponent. (2 + 8) logb (q + t) one year ago one year ago

1. swiftskier96

$2\log_{b}q + 8\log_{b}t$

2. swiftskier96

logb (q^2 + t^8) logb(q^2t^8) logb(qt^(2+8)) In this answer, the 2+8 is the entire exponent. (2 + 8) logb (q + t)

3. swiftskier96

2. 4 log x – 6 log (x + 2) (1 point) 24 log x/x + 2 log x^4(x + 2)^6 log x (x + 2)^24 none of these

4. phi

I don't see what you can do with the first one. for the 2nd one, can't you bring the 4 inside the log using $\log(x^b)= b\log(x)$ in the opposite order same for the 6 then you can replace the subtraction of logs with division inside the log

5. phi

oh, I see what the first question is: make $2\log_{b}q + 8\log_{b}t$ into one log it is the same problem as the other one. bring the 2 inside. bring the 8 inside then change + logs to multiplication inside the log

6. swiftskier96

Is #2 B?

7. phi

B is close, but the -6 is ruining it. you change (you really should try doing this) the expression from 4 log x – 6 log (x + 2) to log x^4 - log( (x+2)^6 ) if we were adding then B would be the answer. but the - means you divide log ( x^4/ (x+2)^6 ) or using negative exponents $\log(x^4 (x+2)^{-6})$ so unless there is a typo, B is not it.

8. androidonyourface

c?

9. swiftskier96

So if you divide the log, it would be A? (Sorry if its wrong. Im really trying my best to understand.)

10. androidonyourface

@phi

11. phi

I don't think the answer is among your choices move the 4 and the 6 inside the log $4 \log x – 6 \log (x + 2)$ becomes $\log( x^4) - \log( (x+2)^6 )$ the - means you divide $\log \left( \frac{x^4}{ (x+2)^6} \right)$

12. swiftskier96

Oh ok. So it would have to be D.

13. phi

yes. can you take a shot at the first problem?

14. swiftskier96

I can try. Hold on.

15. swiftskier96

Would #1 be B?

16. phi

$2\log_{b}q + 8\log_{b}t$ bring the 2 and and 8 inside (all base b, but I'm not typing it in) $\log(q^2) + \log(t^8)$ adding logs changes to multiplication inside a log (in the old days people did this the other way round, because adding is easier then multiplying long numbers) $\log(q^2t^8)$

17. swiftskier96

Hey, i got it right! Lol Thanks!!! :)

18. phi

yw