Difference between revisions of "1991 AHSME Problems/Problem 6"
m (→Solution) |
(Added a solution with explanation) |
||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | <math>\fbox{E}</math> We have <math>\sqrt{x} = x^{\frac{1}{2}}, x\sqrt{x} = x^{\frac{3}{2}}</math>, then square-rooting that gives <math>x^{\frac{3}{4}}</math>, then multiplying by <math>x</math> gives <math>x^{\frac{7}{4}}</math>, then square-rooting this gives <math>x^{\frac{7}{8}}</math>. |
== See also == | == See also == | ||
Revision as of 17:18, 23 February 2018
Problem
If 
, then 
(A) 
(B) ![$x \sqrt[4]{x}$](http://latex.artofproblemsolving.com/7/c/f/7cf418ad893d4471e904ffade1ccdb11447b38b6.png)
(C) ![$\sqrt[8]{x}$](http://latex.artofproblemsolving.com/b/b/e/bbeec62682662d56c5469cae26f1d3fbcdaefce3.png)
(D) ![$\sqrt[8]{x^3}$](http://latex.artofproblemsolving.com/5/d/9/5d988409eaa26897191cc88bcd239de0fe0f2c47.png)
(E) ![$\sqrt[8]{x^7}$](http://latex.artofproblemsolving.com/e/0/d/e0d6e780b78d3b4b8c67a71d9b3b0490b731b3a3.png)
Solution
We have 
, then square-rooting that gives 
, then multiplying by 
gives 
, then square-rooting this gives 
.
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
