Difference between revisions of "1953 AHSME Problems/Problem 22"
(The solution was originally just the solution, so I added everything else and rewrote the solution.) |
m (I added a space.) |
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\textbf{(D)}\ 3 \qquad | \textbf{(D)}\ 3 \qquad | ||
\textbf{(E)}\ \text{none of these} </math> | \textbf{(E)}\ \text{none of these} </math> | ||
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==Solution== | ==Solution== | ||
<math>27\sqrt[4]{9}\sqrt[3]{9}</math> can be rewritten as <math>3^3\cdot 3^\frac{1}{2}\cdot 3^\frac{2}{3}</math>. Using exponent rules, this simplifies to <math>3^\frac{25}{6}</math>. | <math>27\sqrt[4]{9}\sqrt[3]{9}</math> can be rewritten as <math>3^3\cdot 3^\frac{1}{2}\cdot 3^\frac{2}{3}</math>. Using exponent rules, this simplifies to <math>3^\frac{25}{6}</math>. | ||
Latest revision as of 20:30, 27 November 2019
Problem
The logarithm of ![$27\sqrt[4]{9}\sqrt[3]{9}$](http://latex.artofproblemsolving.com/a/a/8/aa8fef328edfb54623da396a77f766175ad4aed8.png)
to the base 
is:
Solution
can be rewritten as 
. Using exponent rules, this simplifies to 
.
The problem wants us to find ![$\log_3{27\sqrt[4]{9}\sqrt[3]{9}}$](http://latex.artofproblemsolving.com/2/b/e/2be9a73777b718376ebf0a3be9606b0e6015be1d.png)
. We just found that this is equal to 
. Using logarithm rules, this is equal to 
, which is simply 
. The answer is 
.
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
