Finding a×b when (log a, log b) is on the unit circle

[dfklajsdfald]

Homework Statement

the point (log a, log b) exists on the unit circle. find the value of axb. round to the nearest thousandths.

Homework Equations

x² + y² = 1

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 wouldn't 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 I am not sure if that's 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

 

[haruspex]

the point (log a, log b) exists on the unit circle. find the value of axb. round to the nearest thousandths.

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.

 

[dfklajsdfald]

nope i just checked the question and what i said is exactly what's 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

 

[haruspex]

nope i just checked the question and what i said is exactly what's 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

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=c^xc^y=c^x+y, so v=x+y. So it is the same as asking for the value of x+y given that x²+y²=1.

 

[Mark44]

Homework Statement

the point (log a, log b) exists on the unit circle. find the value of axb. round to the nearest thousandths.

Homework Equations

x² + y² = 1

The Attempt at a Solution

x2+y2 = 1
loga²+logb² =1
2loga+2logb = 1

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

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 wouldn't 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 I am not sure if that's 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