Question

12^11 = 2^x
How can I use log to solve this problem?

Answer 1

Sorry the site went down and I lost everything I typed. let me try again...

Answer 2

Sad face. Thanks for the effort though

Answer 3

If you have:\[a^b=c^d\]Then you can take logs of both sides to get:\[\log(a^b)=\log(c^d)\]\[\therefore b\log(a)=d\log(c)\]You should be able to use this to solve your problem.

Answer 4

Let me know f you need more help and/or explanation.

Answer 5

what does the 3 dots mean before blog?

Answer 6

\(\therefore\) mean "therefore"

Answer 7

okay, let me see if i can get it now, just a sec

Answer 8

is the answer 39.5 approximately?

Answer 9

perfect! - well done

Answer 10

thanks, but i need to ask you something

Answer 11

go ahead...

Answer 12

the original question is the product of 12^11 x 18^13 = 2^p x 3^q, and it wants me to find 4q - 3p, and all of the possible answers are integers.

Answer 13

ok - interesting problem - it me think a bit...

Answer 14

ok - I think the best approach is try and replace the 12 and 18 by their prime factors.\[12=2^2*3\]\[18=2*3^2\]

Answer 15

can you "see" how to solve it then?

Answer 16

2^2 x 3 x 2 x 3^2 = 2^3 x 3^3!
Then you multiply by 11 and 13 respectively and get p = 33 and q = 39, 4 x 39 - 3 x 33 = 57 which isnt a possible answer. :(, did i mess up somewhere?

Answer 17

not quite - let me take you through it step-by-step...

Answer 18

thanks

Answer 19

\[12^{11}=(12)^{11}=(2^2*3)^{11}=2^{22}*3^{11}\]use same technique with 18 and then see what you get.

Answer 20

I used the rule:\[(x^a)^b=x^{ab}\]

Answer 21

ohh, forgot about that; p = 35, q = 37? So the answer is... 43, and its a possible answer! Yay

Answer 22

well done - you are a fast learning!

Answer 23

*learner

Answer 24

Thanks. You are a good teacher.

Answer 25

:) Thanks

Answer 26

United Kingdom? What's it like there?