Difference between revisions of "1970 AHSME Problems/Problem 3"

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== Solution ==
 
== Solution ==
Since we want the <math>y</math> expression in terms of <math>x</math>, let's convert the <math>y</math> expression. We can convert it to <math>1+ \frac{1}{2p} \Rightarrow \frac{2p+1}{2p} \Rightarrow \frac{x}{x-1} \Rightarrow</math> <math>\fbox{C}</math>
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Since we want the <math>y</math> expression in terms of <math>x</math>, let's convert the <math>y</math> expression. We can convert it to <math>1+ \frac{1}{2^p} \Rightarrow \frac{2^p+1}{2^p} \Rightarrow \frac{x}{x-1} \Rightarrow</math> <math>\fbox{C}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 00:04, 16 January 2019

Problem

If $x=1+2^p$ and $y=1+2^{-p}$, then $y$ in terms of $x$ is

$\text{(A) } \frac{x+1}{x-1}\quad \text{(B) } \frac{x+2}{x-1}\quad \text{(C) } \frac{x}{x-1}\quad \text{(D) } 2-x\quad \text{(E) } \frac{x-1}{x}$

Solution

Since we want the $y$ expression in terms of $x$, let's convert the $y$ expression. We can convert it to $1+ \frac{1}{2^p} \Rightarrow \frac{2^p+1}{2^p} \Rightarrow \frac{x}{x-1} \Rightarrow$ $\fbox{C}$

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

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