can someone help w/ logs

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can someone help w/ logs

Mathematics
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example please
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.
\[3^ (x+1) = 6^x\]

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wait thats wrong what i posted
\[3^{x+1}=6^x\]
its 3 ^ (x+1) = 6^x
yes that
has nothing to do with logs.
oops i lied excuse me
yes it does you have to multiply each side by a log...huh?
no you do not 'multiply' by the log you 'take the log'
you can choose any base you like, but I will use \[log_e\] which is written \[ln\]
thats what i mean...im so confused
take the log of both sides to get \[ln(3^{x+1})=ln(6^x)\]
then what? log and what base? dont u mean log3 (log base 3 i mean)
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.
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...
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.
ok what do i do after (x+1) log (3) = x log 6
both sides... 6=3*2, so you can present log3 (3*2)= 1+log3(2) (log3 - log with base of 3)
ahhh too many people are helping me im so confused
the same as in ln... you'll have ln or log in answer
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 \[2^x=1000\]?
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.
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
so if i want to solve \[2^x=1000\] for x i take the log of both sides and get \[log(2^x)=log(1000\] \[xlog(2)=log(1000)\] \[x=\frac{log(1000)}{log(2)}\]
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 \[x\log(3)+log(3)=xlog(6)\] \[log(3)=xlog(6)-xlog(3)\] \[log(3)=x(log(6)-log(3))\] \[\frac{log(3)}{log(6)-log(3)}=x\]
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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