Question

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

I did:

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

So

(2x-2)^2 = (1-x)^4)
0 = x^2 - 4x + 3
Solutions: 3 and 1

But I got it wrong and the solution is only -1...how?

Answer 1

\[\log((2x-2)^{2}) = 4\log(1 - x)\]

Answer 2

\[\log((2x - 2)^{2}) = \log((1 - x)^{4})\]\[(2x - 2)^{2} = (1 - x)^{4}\]\[\sqrt{(2x - 2)^{2}} = \sqrt{(1 - x)^{4}}\]\[2x - 2 = (1 - x)^{2}\]\[2x - 2 = x^{2} - 2x + 1\]\[0 = x^{2} -4x + 3\]

Answer 3

So the factors are \[0 = (x - 3)(x - 1)\] Which means the solutions are 3, and 1 but those are not correct and the solution is -1

Answer 4

Do you understand the domain and range of the log function?

Answer 5

What is the log of (1-3)?

Answer 6

I know the answers can't be 1 or 3 because it gives negative logs which is impossible, so that means there are no solutions...where did -1 come from?

Answer 7

@failmathmajor

Look again at your solutions and at the equations you have already derived, now explain to me why shouldn't they be correct?

Answer 8

\[\begin{align}
\log((2x-2)^{2}) &= 4\log(1 - x)\\
\log(4(x-1)^2)&=\log((1-x)^4)\\
\log(4)+\log((x-1)^2)&=\log((1-x)^4)\\
\log(4)&=\log((1-x)^4)-\log((x-1)^2)\\
&=\log(\frac{(1-x)^4}{(x-1)^2})\\
&=\log(\frac{(1-x)^4}{(1-x)^2})\\
&=\log((1-x)^2)\\
\therefore 2\log(2)&=2\log(1-x)\\
\therefore 2&=1-x\\
\therefore x&=-1\\
\end{align}\]

Answer 9

sorry it took so long - I lost internet for a while :(

Answer 10

Spacelimbus if you plug in 3 on the right side of the equation, you get \[= 4 \log(1 - 3)\] therefore a negative log so the statement cannot be true. If you plug in 1 for either side, for example the right side, you get: \[4 \log(1 - 1)\] which is the log of 0 which is also undefined. So -1 is the only correct solution

Answer 11

Yes I agree, pardon me I was skipping an important step.

Obviously you need the extra step @asnaseer has already mentioned. I was mislead by the LHS of the equation.

Answer 12

you could also get to this answer from your step as follows:\[\begin{align}
(2x - 2)^{2} &= (1 - x)^{4}\\
4(x-1)^2&=(1-x)^4\\
4(x-1)^2-(1-x)^4&=0\\
4(1-x)^2-(1-x)^4&=0\\
(1-x)^2(4-(1-x)^2)&=0\\
\end{align}\]which leads to:\[x=1\]or:\[1-x=\pm2\]i.e.:\[x=3\text{ or }-1\]so the full solutions are: x=-1, 1 or 3

but 1 and 3 can be disregarded as they don't fit the domain of logs.

Answer 13

you /lost/ the x=-1 solution when you took square roots as you didn't cater for \(\pm\)

Answer 14

Oh, so that was it

Answer 15

I must admit though that this was a tricky problem :)