Question

log(log(x²))+log(log(x-4))=1
What I tried so far:
log(log(x²)*log(x-4))=1
log(x²)*log(x-4)=10
Stuck here, ideas?

Answer 1

log(x²)*log(x-4)=10
2log(x)*log(x-4)=10
log(x)*log(x-4)=5
after that hmm...

Answer 2

Thought of doing that too, but lead me nowhere.
The answer claims to be 10^5, but I don't see how.

Answer 3

It sort of works if I remove part of the problem and make it log(log(x²))=1
And even then it's ±10^5

Answer 4

That is not the correct answer.
If you are allowed to use graphing calculator you can plot log(x)*log(x-4)-5 and see where it crosses the x-axis. If you are not allowed to use a graphing calculator you can solve it using trial and error. The answer is \(x \approx 11.6584\)

Answer 5

Can you post a screenshot of the problem?

Answer 6

http://puu.sh/aZ96p/b285ebe01e.jpg

Answer 7

Then 11.6584 is the correct answer and not 10^5.

Answer 8

I just checked it in the calc function, the log flank equals to 1.698968... and not 1

Answer 9

Seems like an error in the textbook indeed, thanks for verifying it's a mistake, now I can die in peace.

Answer 10

lol. yw.

Answer 11

\(\log(\log x - 4 ) \ne \log(\log(x-4))\)

Answer 12

good catch. glad I asked you to post the screenshot. In reproducing the problem you have posted an extra pair of parenthesis around x-4.

Answer 13

I think you should be able to solve it with a quadratic near the end

Answer 14

log(x²) * { log(x)-4 } =10
2log(x) * { log(x)-4 } =10
log(x) * { log(x)-4 } = 5
{log(x)}^2 - 4log(x) - 5 = 0
Let t = log(x)
t^2 - 4t - 5 = 0
(t-5)(t+1) = 0
t = 5, t = -1
log(x) = 5, log(x) = -1.
x = 10^5, x = 10^(-1) = 0.1. The second solution may be extraneous.

Answer 15

Yes, the second solution is extraneous and can be discarded because when x = 0.1, log(log(x) - 4) in the original problem becomes log of a negative number which is not allowed.

Answer 16

thanks guys