Question

can someone help w/ logs

Answer 1

example please

Answer 2

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.

Answer 3

\[3^ (x+1) = 6^x\]

Answer 4

wait thats wrong what i posted

Answer 5

\[3^{x+1}=6^x\]

Answer 6

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

Answer 7

yes that

Answer 8

has nothing to do with logs.

Answer 9

oops i lied excuse me

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

thats what i mean...im so confused

Answer 14

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

Answer 15

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

Answer 16

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.

Answer 17

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...

Answer 18

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.

Answer 19

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

Answer 20

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

Answer 21

ahhh too many people are helping me im so confused

Answer 22

the same as in ln... you'll have ln or log in answer

Answer 23

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 \[2^x=1000\]?

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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
\[x\log(3)+log(3)=xlog(6)\]
\[log(3)=xlog(6)-xlog(3)\]
\[log(3)=x(log(6)-log(3))\]
\[\frac{log(3)}{log(6)-log(3)}=x\]

Answer 28

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.