Problem: If 60a=360^{a}=360a=3 and 60b=560^{b}=560b=5, then 12[(1−a−b)/2(1−b)]12^{[(1-a-b) / 2(1-b)]}12[(1−a−b)/2(1−b)] is
Answer Choices:
A. 3\sqrt{3}3
B. 222
C. 5\sqrt{5}5
D. 333
E. 12\sqrt{12}12
Solution:
12=605=6060b=601−b12=\dfrac{60}{5}=\dfrac{60}{60^{b}}=60^{1-b}12=560=60b60=601−b. So
12[(1−a−b)2(1−b)]=[601−b](1−a−b)/2(1−b)=60(1−a−b)/2=6060a60b=603⋅5=2.\begin{aligned} 12^{[(1-a-b) 2(1-b)]} & =\left[60^{1-b}\right]{ }^{(1-a-b) / 2(1-b)} \\ & =60^{(1-a-b) / 2}=\sqrt{\dfrac{60}{60^{a} 60^{b}}} \\ & =\sqrt{\dfrac{60}{3 \cdot 5}}=2 . \end{aligned} 12[(1−a−b)2(1−b)]=[601−b](1−a−b)/2(1−b)=60(1−a−b)/2=60a60b60=3⋅560=2.