## anonymous 5 years ago can someone help w/ logs

1. anonymous

2. anonymous

what would you like to know? the main thing to remember is that logs are actually exponents (until you get to fancy math) so all the laws of exponents apply.

3. anonymous

$3^ (x+1) = 6^x$

4. anonymous

wait thats wrong what i posted

5. anonymous

$3^{x+1}=6^x$

6. anonymous

its 3 ^ (x+1) = 6^x

7. anonymous

yes that

8. anonymous

has nothing to do with logs.

9. anonymous

oops i lied excuse me

10. anonymous

yes it does you have to multiply each side by a log...huh?

11. anonymous

no you do not 'multiply' by the log you 'take the log'

12. anonymous

you can choose any base you like, but I will use $log_e$ which is written $ln$

13. anonymous

thats what i mean...im so confused

14. anonymous

take the log of both sides to get $ln(3^{x+1})=ln(6^x)$

15. anonymous

then what? log and what base? dont u mean log3 (log base 3 i mean)

16. anonymous

then the exponents come out as multipliers to give $(x+1)ln(3)=xln(6)$ we solve this equation for x, remembering that $ln(3)$ and $ln(6)$ are just constants.

17. anonymous

you "take" ln and use a ln properties... (x+1)ln3=xln6 (x+1)/x = ln6/ln3 1+ 1/x=ln6/ln3 1/x=ln6/ln3 -1 x=ln3/(ln6-ln3) you can use log with base of 3 as well...

18. anonymous

let me start slowly. You have two different bases here, 3 and 6. whatever you do to one side you must do the same to the other. you cannot take the log base 3 of one side and the log base 6 of the other.

19. anonymous

ok what do i do after (x+1) log (3) = x log 6

20. anonymous

both sides... 6=3*2, so you can present log3 (3*2)= 1+log3(2) (log3 - log with base of 3)

21. anonymous

ahhh too many people are helping me im so confused

22. anonymous

23. anonymous

let me see if i can unconfuse you. how do you solve any equation with the variable in the exponent? for example how would you solve $2^x=1000$?

24. anonymous

you cannot use 'algebra' because the variable is in the sky. it is in the exponent not on the ground floor, so you cannot add or subtract, multiply or divide to solve for x.

25. anonymous

you solve it by taking the log of both sides, because one of the facts of logs is that $log_b(a^x)=xlog_b(a)$ and now the variable is on the ground floor. so can solve using algebra

26. anonymous

so if i want to solve $2^x=1000$ for x i take the log of both sides and get $log(2^x)=log(1000$ $xlog(2)=log(1000)$ $x=\frac{log(1000)}{log(2)}$

27. anonymous

now you have $3^{x+1}=6^x$ tale the log of both sides to get $log(3^{x+1})=log(6^x)$ $(x+1)log(3)=xlog(6)$ multiply out to get $x\log(3)+log(3)=xlog(6)$ $log(3)=xlog(6)-xlog(3)$ $log(3)=x(log(6)-log(3))$ $\frac{log(3)}{log(6)-log(3)}=x$

28. anonymous

which base is the log? it doesn't matter. but if you want a number out of this you have to use a base that you can find on your calculator. you can use log base ten which is just written as log or you can use log base e which is written as ln.

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