Difference between revisions of "1983 AHSME Problems/Problem 25"
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==Problem 25== | ==Problem 25== | ||
| − | If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^ | + | If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^{(1-a-b)/\left(2\left(1-b\right)\right)}</math> is |
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad</math> | <math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad</math> | ||
==Solution== | ==Solution== | ||
| − | + | We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get | |
| + | <cmath>12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.</cmath> | ||
| + | Hence | ||
| + | <cmath>12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.</cmath> | ||
| + | Since <math>4=\frac{60}{3\cdot 5}</math>, we have | ||
| + | <cmath>4=\frac{60}{60^a 60^b}=60^{1-a-b}.</cmath> | ||
| + | Therefore, we have | ||
| + | <cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath> | ||
| − | |||
| − | + | == Solution 2 == | |
| + | We have <math>60^a = 3</math> and <math>60^b = 5</math>. We can say that <math>a = \log_{60} 3</math> and <math>b = \log_{60} 5</math>. | ||
| − | + | <cmath>12^{(1-a-b)/[2(1-b)]} = 12^{(1-(a+b))/[2(1-b)]}</cmath> | |
| − | <math> | + | We can evaluate (a+b) by the Addition Identity for Logarithms, <math>(a+b) = \log_{60} 15</math>. Also, <math>1 = \log_{60} 60</math>. |
| − | < | + | <cmath> (1-(a+b) = \log_{60} 60 - \log_{60} 15 = \log_{60} 4 </cmath> |
| − | + | Now we have to evaluate [2(1-b)], the denominator of the fractional exponent. Once again, we can say <math>1 = \log_{60} 60</math> | |
| + | |||
| + | <cmath> 2(1-b) = 2(\log_{60} 12)</cmath> | ||
| + | |||
| + | <cmath>12^{(\log_{60} 4)/[2(\log_{60} 12]} = 12^{\frac{1}{2} \cdot \log_{12} 4} = 4^{1/2} = 2</cmath> | ||
| + | |||
| + | ~YBSuburbanTea | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1983|num-b=24|num-a=26}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 11:12, 14 January 2022
Contents
Problem 25
If 
and 
, then 
is
Solution
We have that 
. We can substitute our value for 5, to get
![\[12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.\]](http://latex.artofproblemsolving.com/3/e/9/3e9f343c772c6fcabb7c15c2b8e0ec5e56dfc51b.png)
Hence
![\[12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.\]](http://latex.artofproblemsolving.com/6/e/b/6ebb5544f77ce86e1d129a0a929377b0a3a8f9e1.png)
Since 
, we have
![\[4=\frac{60}{60^a 60^b}=60^{1-a-b}.\]](http://latex.artofproblemsolving.com/d/2/f/d2f41a20a20ce8e7f41c6519afb08e70677589f6.png)
Therefore, we have
![\[60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}\]](http://latex.artofproblemsolving.com/2/6/9/26995b76464638b54f8741cf13123f3c606016a4.png)
Solution 2
We have 
and 
. We can say that 
and 
.
![\[12^{(1-a-b)/[2(1-b)]} = 12^{(1-(a+b))/[2(1-b)]}\]](http://latex.artofproblemsolving.com/0/7/8/078b538042a748d409b847d511137a3bbb8e0359.png)
We can evaluate (a+b) by the Addition Identity for Logarithms, 
. Also, 
.
![\[(1-(a+b) = \log_{60} 60 - \log_{60} 15 = \log_{60} 4\]](http://latex.artofproblemsolving.com/e/a/e/eaeb9e65a0de20a302e8708326482a4bf65e4eee.png)
Now we have to evaluate [2(1-b)], the denominator of the fractional exponent. Once again, we can say
![\[2(1-b) = 2(\log_{60} 12)\]](http://latex.artofproblemsolving.com/b/3/9/b395fd50a35b30a2cad1cc082e521e4109b2f50e.png)
![\[12^{(\log_{60} 4)/[2(\log_{60} 12]} = 12^{\frac{1}{2} \cdot \log_{12} 4} = 4^{1/2} = 2\]](http://latex.artofproblemsolving.com/2/0/5/2055d9c657a7ac845cba5bc4c3f0b1497da1f09b.png)
~YBSuburbanTea
See Also
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
