Difference between revisions of "1966 AHSME Problems/Problem 12"
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== Solution == | == Solution == | ||
We know that | We know that | ||
| − | <math>2^{6x+3} | + | <math>2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}</math>. |
We also know that | We also know that | ||
<math>8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}</math>. | <math>8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}</math>. | ||
| − | + | There are infinite solutions to the equation <math>2^{12x+15}=2^{12x+15}</math>, so the answer is <math>\boxed{\text{(E) greater than 3}}</math>. | |
== See also == | == See also == | ||
Latest revision as of 18:48, 24 March 2023
Problem
The number of real values of 
that satisfy the equation ![\[(2^{6x+3})(4^{3x+6})=8^{4x+5}\]](http://latex.artofproblemsolving.com/b/7/5/b753e8412922d2d1b09296d555439987d31f2308.png)
is:
Solution
We know that
.
We also know that
.
There are infinite solutions to the equation 
, so the answer is 
.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
