# Point on a unit circle

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1. Dec 2, 2016

### dfklajsdfald

1. The problem statement, all variables and given/known data
the point (log a, log b) exists on the unit circle. find the value of axb. round to the nearest thousandths.

2. Relevant equations
x2 + y2 = 1

3. The attempt at a solution

x2+y2 = 1
loga2+logb2 =1
2loga+2logb = 1
2(loga+logb) = 1
loga + log b = 0.5
logb = 0.5−loga
now i try and subsitute logb in
loga2+(0.5−loga)2 = 1

when i did this it wouldnt work after the last step. so this is what i tried next

(loga)2 + (log b)2 = 1
loga = √1-(logb)2

then i did

(√1-(logb)2)2 + (logb)2 = 1
1-(logb)2 = 1-(log b)4
-(logb)2 + (logb)4 = 0
factored out (logb)2 so i got
1 = logb
and 101 = b so b = 10 but im not sure if thats right either because it seems iffy to me
i think i was on the right track with the first one but idk can someone help please

2. Dec 2, 2016

### haruspex

That makes no sense. There are infinitely many solutions.
E.g. a=1, b=e gives e;
ab=1 gives 1;
a=b gives e√2.
Please check you have stated the question exactly.

3. Dec 2, 2016

### dfklajsdfald

nope i just checked the question and what i said is exactly whats written. it says: the point (log a, log b) exists on the unit circle. find the value of a times b. round to the nearest thousandth

4. Dec 3, 2016

### haruspex

I just noticed you were assuming log base 10, whereas I assumed natural logs, but my answer applies whatever the base. It is more obvious if we substitute x=log a, y= log b, v=log (ab). Let the base be c. ab=cxcy=cx+y, so v=x+y. So it is the same as asking for the value of x+y given that x2+y2=1.

5. Dec 3, 2016

### Staff: Mentor

No.
The second equation is $(\log(a))^2 + (\log(b))^2 = 1$. This is not the same as $\log(a^2) + \log(b)^2 = 1$.