anonymous
  • anonymous
Which of the following shows the extraneous solution to the logarithmic equation below? log3(18x^3)-log3(2x)=log3(144) x=-16 x=-8 x=-4 x=-2 I know C is not the extraneous solution, because that is the valid solution
Mathematics
schrodinger
  • schrodinger
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campbell_st
  • campbell_st
well doesn't extraneous mean that it maybe a solution but when you check it, the solution doesn't exist... so substitute you solution into the original equation.... you need to remember that you can't take the log of a negative number....
campbell_st
  • campbell_st
so substitute your choice for a valid solution into the original equation and see if it works..
anonymous
  • anonymous
Can you maybe work out the full problem for me? Because I'm not really sure of the other solution options other than 4 and -4.

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campbell_st
  • campbell_st
ok... x = -4 as a solution \[\log_{3}[18 \times (-4)^3] - \log_{3}[2 \times (-4)] = \] what is the value of the numbers inside the square brackets..?
campbell_st
  • campbell_st
one of the fundemental rules for working with logs is that you can't take the log of a negative number..... so think about that as you calculate the values.. of \[18 \times (-4)^3 ~~and~~ 2 \times (-4) \]
anonymous
  • anonymous
-1152 and -8. So would the answer to the problem be -8?
campbell_st
  • campbell_st
no... the fact is now you are looking at \[\log_{3}(-1152) - \log_{3}(-8)\] and as I said previously you can't take the log of a negative number... so while x = -4 is a solution to the equation when simplified, the fact that when you substitute it into the original equation you end up with \[\log_{3}(-1152) - \log_{3}(-8) = \log_{3}(144)\] it doesn't work... as an example calculate ln(-5) what do you get...?
campbell_st
  • campbell_st
when you put the positive solution in x = 4 you get \[\log_{3}(1152) - \log_{3}(8) = \log_{3}(144)\] which is correct...
anonymous
  • anonymous
So what would the answer be then, since 8 or -8 does not work?
campbell_st
  • campbell_st
the extraneous solution is x = -4 since you can't take the log of a negative number..... as I've said previously. x = 4 is a valid solution....

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