<p>If n is a positive integer and 2^n + 2^(n+1) = k, what is 2^(n+2) in terms of k?</p>
<p>I got to k = 2^n + 2^n x 2^1
but not quite sure how they get to 2^n(1+2) = k
I know take common 2^n out but why +2? is it because the 2^1?</p>
<p>ps I've already looked up the solution but still slightly unclear </p>
<p>and thanks :)</p>
<p>the ans. is 4k/3 ?</p>
<p>Yes. But how do you get 2^n(1+2) = k? why (1+2)?</p>
<p>2^n + 2^(n+1) = k</p>
<p>take the common factor in the left side, which is 2^n. so,
2^n (1+2) = k</p>
<p>2^n (3) = k</p>
<p>2^n = k/3</p>
<p>multiply the 2 sides by 2^2, or 4, then</p>
<p>2^(n+2) = 4k/3</p>
<p>I know I am not very good in explaining, if u didn’t understand I will try another way</p>
<p>Oh but why +2? for this part 2^n (1+2) = k</p>
<p>if it was 2^n + 2^(n+6) = k, would it become 2^n(1+7) = k ? Is my understanding correct?</p>
<p>also if you have time, please show me another way. I’m having a bit of difficulty with these things 
Thanks</p>
<p>2^n + 2^(n+1) = k</p>
<p>Factor 2^n
2^n + (2^n)(2^1) = k</p>
<p>2^n(1+2^1) = k </p>
<p>or 2^n(3) = k </p>
<p>If it was 2^n + 2^(n+6) = k, then again we factor 2^n:</p>
<p>2^n(1 + 2^6) = k </p>
<p>or 2^n(65) = k</p>
<p>I believe that this is actually a multiple choice question. If this is the case I would recommend picking numbers for this problem. Only students very strong in algebra will be able to solve this algebraically on the actual SAT.</p>
<p>
</p>
<p>I would add that students very strong in algebra would be able to solve this problem faster and with more certitude following your advice and forgetting the algebra.</p>
<p>Set n equal to two or three numbers and see what relationship unfolds. We’re looking for 2^(n+2), which I’ll call x because I’m lazy…</p>
<p>*when n=1
k=6 and x=8</p>
<p>when n=2
k=12 and x=16</p>
<p>when n=3
k=24 and x=32</p>
<p>It’s pretty obvious what x equals in terms of k just looking at it and this takes a matter of seconds. Even very strong math students should favor this method over algebra.</p>
<ul>
<li>If you are running through a few numbers, using 1 is fine. Never use 0 or 1 in isolation though when using this kind of pick-a-number strategy.</li>
</ul>
<p>Thank you for all your replies 
It’s a lot clearer than before</p>
<p>Is my thinking correct?
2^n + 2^(n+6) = k
2^n (1+(1x2^6)) = k</p>
<p>?</p>
<p>Yes 10char</p>
<p>primalzerg,</p>
<p>yes, that is the correct way to factor. </p>
<p>here is another problem from the official SAT(easier):</p>
<p>If k is a positive integer, which of the following is
equivalent to (3^k) + (3^k) ?
(A) 2(3^k)
(B) 3^(2k)
(C) 6^k
(D) 6^(2k)
(E) 9^(2k)</p>
<p>Is the answer B?</p>
<p>The easiest way to solve this questions was to put n as 0 and then look for the answer in the choices</p>
<p>primalzerg,</p>
<p>no it is not B. i would recommend trying some simple numbers for k and testing the answer choices. after that try factoring in the same fashion as you did earlier. </p>
<p>in the exponent problems, there is no general rule for addition. the common temptation i see with students is to take 5^x + 5^x and equate it to 5^(2x). the best bet is to always factor, or use simple numbers.</p>
<p>For your initial question. When we reach 2^n=k/3. We need to realize how we can turn the left hand side into 2^(n+2). Which is why we multiply 4 to it. However we need to do to the right hand the same we do to the left hand to keep equation balanced. So we get 2^(n+2) = 4k/3. For the second question posed, consider 3^x = y. What is y+y? 2y. Now substitute 3^x for y. We get 2(3^x). I agree that strong algebra is required. So plugging in numbers can actually help in ALOT of the sat questions.</p>
<p>3^k* i thought it was x. Nevertheless, same concept</p>
<p>the answer for the second question is A</p>
<p>Oh, yes it is A. I strongly dislike algebra. Anyone got a website that provides this type of questions? I want to be more comfortable with this.</p>
<p>Thank you for all your replies :)</p>
<p>no it is not B, factor it out</p>