anonymous
  • anonymous
can someone help w/ logs
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
example please
anonymous
  • 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.
anonymous
  • anonymous
\[3^ (x+1) = 6^x\]

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