<p>X: -1, 0 , 1
f(x): (1/8), (1/2), 2</p>
<p>The table above shows values for the function f. If F(x)= ka^x for some constants k and a, what is the value of a?</p>
<p>a. (1/2)
b. (1/4)
c. 2
d. 4
e. 16</p>
<p>Correct Answer: D</p>
<p>k*a^0=1/2—>k=1/2</p>
<p>1/2/a=1/8
1/a=1/4
a=4</p>
<p>OR you could use x=1</p>
<p>ka=2
a/2=2
a=4</p>
<p>Gots it?</p>
<p>f(x) = ka^x
Use the value 1/2 for f(x), and 0, the x value that gives 1/2 for f(x)
1/2 = ka^0
Notice a^0 is just 1 (any number raised to 0 is 1)
1/2 = k(1)
1/2 = k
We now have k = 1/2</p>
<p>Now use the number 2 for f(x), and you can see in the table that the x value 1 gives this
f(x) = ka^x
2 = ka^1
2 = (1/2)a^1
a^1 is just a
2 = (1/2)a
Multiply both sides of the equation by 2 to get a alone
4 = a</p>
<p>^ Very well explained.</p>
<p>My feelings are hurt :-P</p>
<p>Yours was well explained too, but just wasn’t clear as how the other guy explained it.</p>
<p>Maybe you can try to reason it for us better next time, there will always be another question coming up on the SAT forum.</p>
<p>I was just teasing you. I had just planned on working it there and explaining it, but I accidentally hit submit before I could add explanations lol.</p>
<p>Well to be honest, both ways that were started out, wasn’t greatly detailed by the procedure method though. </p>
<p>You should try writing out the entire question, show each variable one by one and lead us to how the answer was revealed. Don’t just use some random variables to make them into questions. Lol</p>
<p>Darn it, that just screws my next explanation:
y=x^x
ln y= xln x
y’/y=ln x +1
y’=x^x(ln x +1)</p>
<p>Ok, this thread is dead.</p>
<p>^not quite. :)</p>
<p>ka^1 / ka^0 = 2 / (1/2)
a = 4</p>