# Thread: System of Equations, have to use exponent laws

1. ## System of Equations, have to use exponent laws

60^a=3 and 60^b=5. I have to find out the value of 12^((1-a-b)/(2-2b))? All i have figured out is 60^1-a-b=4 because 60^a+b=3*5=60/4, so 60^a+b-1=1/4, so 60^1-a-b=4. Don't know if it is relevant for this problem

2. ## Re: System of Equations, have to use exponent laws

Originally Posted by Ilikebugs
60^a=3 and 60^b=5.
I have to find out the value of 12^((1-a-b)/(2-2b)).
All i have figured out is 60^1-a-b=4 because 60^a+b=3*5=60/4,
so 60^a+b-1=1/4, so 60^1-a-b=4.
Can't tell what you're doing; you're showing sloppy stuff,
like 60^1-a-b which should be 60^(1-a-b).

Anyway:
60^a = 3 ; so a = log(3) / log(60); similarly b = log(5) / log(60)

Substitute results in 12^[(1-a-b)/(2-2b)] to get solution.

3. ## Re: System of Equations, have to use exponent laws

$60=2^2 3^1 5^1$

rewrite the two equations

$2^{2a}3^{a-1}5^a=1$

$2^{2b}3^b5^{b-1}=1$

now eliminate $5$ (since the answer involves only the primes $2$ and $3$)

...

$2^{2a}=3^{1-a-b}$

next eliminate $3$