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

 
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<math>0 = y - 6</math>
 
<math>0 = y - 6</math>
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<math>y = 6</math>
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 +
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>
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==See Also==
 
==See Also==
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{{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>
 

Latest revision as of 11:33, 23 July 2019

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 $\boxed{\textbf{(D)}}$


See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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