Difference between revisions of "1959 AHSME Problems/Problem 5"
(Created page with "== Problem 5== The value of <math>\left(256\right)^{.16}\left(256\right)^{.09}</math> is: <math>\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ ...") |
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==Solution== | ==Solution== | ||
| − | When we multiply numbers with exponents, we add | + | When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate <math>256^{0.16} \cdot 256^{0.09}</math>: |
<cmath>256^{0.16} \cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.</cmath> | <cmath>256^{0.16} \cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.</cmath> | ||
Now we can convert the decimal exponent to a fraction: <cmath>256^{0.25} = 256^{\frac{1}{4}}.</cmath> | Now we can convert the decimal exponent to a fraction: <cmath>256^{0.25} = 256^{\frac{1}{4}}.</cmath> | ||
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Since <math>256 =16^2=(4^2)^2=4^4</math>, we can solve for the fourth root of <math>256</math>: <cmath>\sqrt[4]{256}=\sqrt[4]{4^4}=4.</cmath> | Since <math>256 =16^2=(4^2)^2=4^4</math>, we can solve for the fourth root of <math>256</math>: <cmath>\sqrt[4]{256}=\sqrt[4]{4^4}=4.</cmath> | ||
Therefore, <math>(256)^{.16} \cdot (256)^{.09}=\boxed{\textbf{(A) }\ 4}.</math> | Therefore, <math>(256)^{.16} \cdot (256)^{.09}=\boxed{\textbf{(A) }\ 4}.</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME 50p box|year=1959|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:08, 25 February 2018
Problem 5
The value of 
is:
Solution
When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate 
:
![\[256^{0.16} \cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\]](http://latex.artofproblemsolving.com/3/1/3/31384237d46ca82b0154788ab73b9df0bc2c89c5.png)
Now we can convert the decimal exponent to a fraction: ![\[256^{0.25} = 256^{\frac{1}{4}}.\]](http://latex.artofproblemsolving.com/c/1/d/c1d16907bb57ffe5cd7691716308a373307be0c6.png)
Now, let us convert the expression into radical form. Since 
is the denominator of the fractional exponent, it will be the index exponent: ![\[256^{\frac{1}{4}}=\sqrt[4]{256}.\]](http://latex.artofproblemsolving.com/c/1/f/c1f0d3747b9ac71a5fcace8ede61809f935fa386.png)
Since 
, we can solve for the fourth root of 
: ![\[\sqrt[4]{256}=\sqrt[4]{4^4}=4.\]](http://latex.artofproblemsolving.com/6/f/8/6f8b2a71bcd4bb50876530774b5c9972f0190ab9.png)
Therefore, 
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
