<p>If n is a positive integer and (n+1)/(2^n) = 1/2, then n = </p>
<p>A) 1
B) 2
C) 3
D) 4
E) 5</p>
<p>The answer is easy to find. It's C. How can you solve this algebraically?</p>
<p>It seems like the easiest ways by far are guessing and checking the answers/intuition, or simply using the intersection function on a graphing calculator. </p>
<p>Otherwise, as far as I know, it’d be VERY complex for SAT standards.</p>
<p>This is Test 9 (7 if you have the older one) Section 5 #6.</p>
<p>This is probably the most difficult problem I’ve ever encountered on the SAT.</p>
<p>Perhaps you’re supposed to solve it through some sort of mix of intuition and some rule/law, but they can’t expect you to actually solve for N a normal way.</p>
<p>I’ll think of it some more.</p>
<p>I’m still thinking.</p>
<p>I’m here and stuck:
n+1=2^(n-1)</p>
<p>The solution for 3 is:
<a href=“http://www4a.wolframalpha.com/Calculate/MSP/MSP37919hedb4a4ec47d79000021bbedbh7083g6ic?MSPStoreType=image/gif&s=26&w=206&h=50[/img]”>http://www4a.wolframalpha.com/Calculate/MSP/MSP37919hedb4a4ec47d79000021bbedbh7083g6ic?MSPStoreType=image/gif&s=26&w=206&h=50
</a></p>
<p>So as I said, I don’t know how they expect you to solve it other than checking possible integers.</p>
<p>Aren’t logarithms supposed to be NOT included on the SAT ??</p>
<p>I can’t think of a way to solve for an unknown power algebraically…</p>
<p>You’d have to solve it using logarithms, otherwise solve it using logic. </p>
<p>Yes, logarithms aren’t included on the SAT and that’s why it is able to be solved using logic. </p>
<p>There are some questions on the SAT which can be solved using calculus, however calculus isn’t included on the SAT. There are merely problems which CAN BE but DON’T HAVE TO BE solved using calculus. The same holds true with logarithms.</p>
<p>I don’t think you can solve it with even rudimentary knowledge of logarithms.</p>
<p>I’m currently in AP Calc, and I can’t even solve it with AP Calc level math.</p>
<p>Plugging in the answers, starting with c) would be the fastest way for getting the right answer.</p>
<p>However, here is another approach using algebra. Using this approach, we still need to try values for n, one-by-one. However, you will see exactly why the quotient is 1/2. </p>
<p>By the binomial theorem (see wikipedia if you haven’t heard of this theorem)</p>
<p>2^1 = (1+1) = 1 n+1 = 1+1
2^2 = (1+1)^2 = 1 + 2 + 1 ;n+1 = 2+1
2^3 = (1+1)^3 = 1 + 3 + 3 + 1 ;n+1 = 3+1
2^4 = (1+1)^4 = 1 + 4 + 6 + 4 + 1 ;n+1 = 4+1
2^5 = (1+1)^5 = 1+5+ 10+ 10+ 5 +1 ;n+1 = 5+1</p>
<p>For n = 3 , (n+1)/2^3 = (3+1) / (1+3+3+1) = 1/2</p>
<p>Of course, that is still not an analytic solution, if that is what you have mind.</p>
<p>The graphing calculator approach (numerical solution) would be to first cross multiply the original equation.
2(n+1) = 2^n
Then graph both Y1 = 2(X+1) and Y2 = 2^X and find where the two graphs intersect. They should intersect at roughly X = 4.</p>
<p>You can solve this using algebra,but you have to use logs anyway here is a simplification form.</p>
<p>(n+1)/(2^n) = 1/2 </p>
<p>is same as</p>
<p>(n + 1) * (1/(2^n)) = 1/2;</p>
<p>n + 1 / 2^n = 1/2;</p>
<p>n + 1 = 1/2(2^n);</p>
<p>n + 1 = 2^(-1 + n);</p>
<p>So if you notice you will see that the only real value that suffice the above equation is 3;</p>
<p>
Oops, I meant X = 3.</p>
<p>I don’t think you can solve this analytically (with logs or not). This is a rare problem in that the CB expected you to try different integers (namely, the answers), and that makes it a fairly easy problem. Normally these problems also have algebraic solutions. Not sure this one appeared on an actual SAT, though.</p>