Difference between revisions of "1968 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | |
| + | Because <math>64=4^3</math> and <math>256=4^4</math>, we can change all of the numbers in the equation to exponents with base <math>4</math> and solve the equation: | ||
| + | \begin{align*} | ||
| + | \frac{64^{x-1}}{4^{x-1}}=256^{2x} \\ | ||
| + | \frac{(4^3)^{x-1}}{4^{x-1}}=(4^4)^{2x} \\ | ||
| + | \frac{4^{3x-3}}{4^{x-1}}=4^{8x} \\ | ||
| + | 4^{2x-2}=4^{8x} \\ | ||
| + | 2x-2=8x \\ | ||
| + | x-1=4x \\ | ||
| + | 3x=-1 \\ | ||
| + | x=\frac{-1}{3} \\ | ||
| + | \end{align*} | ||
| + | |||
| + | Thus, our desired answer is <math>\fbox{(B) -1/3}</math>. | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=1|num-a=3}} | + | {{AHSME 35p box|year=1968|num-b=1|num-a=3}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:30, 17 July 2024
Problem
The real value of 
such that 
divided by 
equals 
is:
Solution
Because 
and 
, we can change all of the numbers in the equation to exponents with base 
and solve the equation:
\begin{align*}
\frac{64^{x-1}}{4^{x-1}}=256^{2x} \\
\frac{(4^3)^{x-1}}{4^{x-1}}=(4^4)^{2x} \\
\frac{4^{3x-3}}{4^{x-1}}=4^{8x} \\
4^{2x-2}=4^{8x} \\
2x-2=8x \\
x-1=4x \\
3x=-1 \\
x=\frac{-1}{3} \\
\end{align*}
Thus, our desired answer is 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
