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- 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

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- anonymous

- schrodinger

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- 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

so substitute your choice for a valid solution into the original equation and see if it works..

- 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

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

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

-1152 and -8. So would the answer to the problem be -8?

- 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

when you put the positive solution in x = 4 you get
\[\log_{3}(1152) - \log_{3}(8) = \log_{3}(144)\]
which is correct...

- anonymous

So what would the answer be then, since 8 or -8 does not work?

- 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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