Difference between revisions of "1964 AHSME Problems/Problem 11"

Latest revision as of 05:07, 23 February 2023

Problem

Given $2^x=8^{y+1}$ and $9^y=3^{x-9}$ , find the value of $x+y$

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$

Solution

Since $8^{y + 1} = 2^{3(y+1)}$ and $9^y = 3^{2y}$ , we have:

$2^x = 2^{3(y+1)}$ and $3^{2y} = 3^{x - 9}$

Note that if $a^b = a^c$ , then $b=c$ . Setting the exponents equal gives $x = 3y + 3$ and $2y = x - 9$ . Plugging the first equation into the second equation gives:

$2y = (3y + 3) - 9$

$2y = 3y - 6$

$0 = y - 6$

$y = 6$

Plugging that back in to $x = 3y + 3$ gives $x = 3(6) + 3$ , or $x = 21$ . Thus, $x+y = 21 + 6$ , or $x+y=27$ , which is option $\boxed{\textbf{(D)}}$

1964 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

{{MAA Notice}

@@ Line 23: / Line 23: @@
 <math>0 = y - 6</math>
+<math>y = 6</math>
+Plugging that back in to <math>x = 3y + 3</math> gives <math>x = 3(6) + 3</math>, or <math>x = 21</math>.  Thus, <math>x+y = 21 + 6</math>, or <math>x+y=27</math>, which is option <math>\boxed{\textbf{(D)}}</math>
 ==See Also==
@@ Line 29: / Line 34: @@
 [[Category:Introductory Algebra Problems]]
-{{MAA Notice}}
+{{MAA Notice}
-<math>y = 6</math>
-Plugging that back in to <math>x = 3y + 3</math> gives <math>x = 3(6) + 3</math>, or <math>x = 21</math>.  Thus, <math>x+y = 21 + 6</math>, or <math>x+y=27</math>, which is option <math>\boxed{\textbf{(D)}}</math>

Page

Toolbox

Search

Difference between revisions of "1964 AHSME Problems/Problem 11"

Latest revision as of 05:07, 23 February 2023

Problem

Solution

See Also