Log question (1 Viewer)

[b00m]

This question appeared in my 2 unit half-yearly.. and has since left me with a bit of a conundrum.

Typed it up in paint:

After applying your log laws and solving with algebra, your answers are x=5 and x=-1

but in this case, is the answer " x= -1" invalid?

If you test both answers, x=5 works fine, but x=-1 does not.... unless you move the '2' behind 'x' to make x². But is that a valid reason/explanation to obtain x=-1 as an answer?

I argued that you couldn't move the 2 behind the x as i thought that that would be incorrectly manipulating the question. A friend of mine, on the other hand, thought that both ansers (-1 and 5) were correct.

It's not a difficult question, i'm just unsure?

 

[Trebla]

The solutions you obtain must satisfy the ORIGINAL equation. So here x = -1 is invalid because it does not satisfy the original equation. When you learnt the identity log x² = 2log x, it should have been emphasised that this only holds for x > 0. So when you convert 2log x to log x² you actually pick up the restriction x > 0 which you have to keep in mind when solving.

 

[b00m]

The solutions you obtain must satisfy the ORIGINAL equation. So here x = -1 is invalid because it does not satisfy the original equation. When you learnt the identity log x² = 2log x, it should have been emphasised that this only holds for x > 0. So when you convert 2log x to log x² you actually pick up the restriction x > 0 which you have to keep in mind when solving.

woohoo

Yer that's what i though, it has to satisfy the original equation.

cheers for that.