Maths question

[LoveToStack]

This isn't actually homework, I'm just wondering where I'm going wrong here (there's a question at the end of all the equations).

Q: log(2x-2)² = 4log(1-x)

From there I did this:

2log(2x-2) = 4log(1-x)
log(2x-2) = 2log(1-x)
log(2x-2) = log(1-x)²

Then I equated both sides to get:

2x-2 = (1-x)²
2x-2 = x²-2x+1
x²-4x+3 = 0
(x-1)(x-3) = 0

From there it's safe to assume that x=1, x=3 } however x !=1 because subbing x=1 into gives log(0) which can't be taken. So from there we can assume x=3.

However, just looking at the equation in the question, I can see that x must also equal -1 because:

Subbing for x=3:

log(6-2)² = log(1-3)⁴
log(16) = log(16)

But then subbing for x=-1 also gives:

log(-2-2)² = log(1+1)⁴
log(16) = log(16)

So then I had myself wondering how you're meant to solve the original equation to give both x=-1 and x=3. Taking a longer way round to the answer I decided to leave the brackets as they were and equate them from the outset using binomial expansion for (1-x)⁴:

(2x-2)² = (1-x)⁴
4x-8x+4 = x⁴-4x³+6x²-4x+1
x⁴-4x³+2x²+4x-3 = 0

Then factorizing that as a whole gives me:

(x-1)(x-1)(x+1)(x-3) = 0

From which I can gather that x = 1, -1, 3 where I can then rule out x=1 as an answer but also know about x=-1 as a possible value for x.

So my question is, how can I solve the original equation normally (as in using the rules of logs and powers etc) so that I'm presented with all 3 solutions of x rather than just the two I got when I tried to simplify the logs? I can also see that in my first attempt to solve x I've taken away the positive powers (2&4) which serve to include x=-1 as a solution for x so I understand, kind of, why it doesn't show up as a solution there. But I don't know how to get it using an easier method than my second way. Cheers for any help provided!

[kk.]

ive just quickly used my calculator and you cant do it as youll be taking a log of -2, which isnt possible. ill have another look, but the correct answer is just -1 apparently

---------- Post added 16-09-2010 at 11:29 PM ----------

wait, is the first part, all of it squared, or just the part in brackets? so log((2x-2)^2) or (log(2x-2))^2. If its the latter, i doubt the brackets would be there :S

[kk.]

Q: log(2x-2)² = 4log(1-x)

log((2x-2)²) - log((1-x)⁴) = 0

log [{(2x-2)²} / {(1-x)⁴}] = 0


since log(1)=0

{(2x-2)²} / {(1-x)⁴} = 1

(2x-2)² = (1-x)⁴

4x² - 8x + 4 = x⁴ - 4x³ + 6x² - 4x + 1


x⁴ - 4x³ + 2x² + 4x - 3 = 0


x = 1. x = -1. x = 3

what you did was right. But i think this is the only way you can do it. Youd then put these numbers into the original formula and see which ones work. the only one that will is -1 apparently, according to wolframalpha.

Try TSR for more help?

[Jesus-Egg]

Q: log(2x-2)2 = 4log(1-x)

From there I did this:

2log(2x-2) = 4log(1-x)
log(2x-2) = 2log(1-x)
log(2x-2) = log(1-x)2

When you cancel out the factor of 2 you are getting rid of 2 solutions.

You need to solve
(2x-2)^2 = (1-x)^4

edit: sorry, didn't read the whole post

If you want to get the full solution you have to solve the quartic polynomial

[LoveToStack]

Q: log(2x-2)² = 4log(1-x)

log((2x-2)²) - log((1-x)⁴) = 0

log [{(2x-2)²} / {(1-x)⁴}] = 0


since log(1)=0

{(2x-2)²} / {(1-x)⁴} = 1

(2x-2)² = (1-x)⁴

4x² - 8x + 4 = x⁴ - 4x³ + 6x² - 4x + 1


x⁴ - 4x³ + 2x² + 4x - 3 = 0


x = 1. x = -1. x = 3

what you did was right. But i think this is the only way you can do it. Youd then put these numbers into the original formula and see which ones work. the only one that will is -1 apparently, according to wolframalpha.

Try TSR for more help?

When you cancel out the factor of 2 you are getting rid of 2 solutions.

You need to solve
(2x-2)^2 = (1-x)^4

edit: sorry, didn't read the whole post

If you want to get the full solution you have to solve the quartic polynomial

Thanks for both the replies. I understand that simplifying removes one of the answers so I see why expanding the entire thing is necessary for both values of x. Good to know! Also, wolfram confirms x as -1 and 3.