## anonymous one year ago 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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1. 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....

2. campbell_st

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

3. 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.

4. 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..?

5. 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 $18 \times (-4)^3 ~~and~~ 2 \times (-4)$

6. anonymous

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

7. 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...?

8. 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...

9. anonymous

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

10. 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....