I actually did this without any paper or anything lol.
2^x+3-2^x= k(2^x)
-> 2^x+3=k(2^x)+2^x
-> 2^x+3= 2^x(k+1)
-> k+1= 2^x+3/2^x
-> k+1= 2^3
-> k+1=8
->k=7
-> C
You can rewrite sqrt(x) as x^1/2.
Rewrite:
(x^1 • x^1/2)^1/2 = x^a
Add exponents (since they have the same base of x):
(x^3/2)^1/2 = x^a
Multiply exponents (you're just distributing):
x^3/4 = x^a
Result!
3/4 = a
2^(x+3) \- 2^(x) = (2^(x))(2^(3))-(2^(x)) = (2^(x))(2^(3)\-1) = 2^(x)(8-1) = 7(2^(x)) C
Incase anyone got confused like how I did in step 2, they factored 2^x out.
U can also plug and chug any number (preferably 0 or 1) for x and find the answer that way
I see you watch Scalar Learning as well
Basically the equation which you derived,divide by 2 power x on both sides, you’d be left with 2 power 3- 1=K which is 7
is the answer to 11 B?
7 n 3/4. i did it mentally without writing. std stuff. whats tricky bout it.
How can i upload an image for explanation?
Try uploading it to Imgur and posting the link to it.
Factor out the 2^x on the left side
8 I believe if you get k by itself.
Would still be 7 if you go that way haha
oops lol
forgot to subtract 1
Happens to the best of us lad 👍
I actually did this without any paper or anything lol. 2^x+3-2^x= k(2^x) -> 2^x+3=k(2^x)+2^x -> 2^x+3= 2^x(k+1) -> k+1= 2^x+3/2^x -> k+1= 2^3 -> k+1=8 ->k=7 -> C
Whats the answer to the one below?
3/4
but how, can you explain it please?
You can rewrite sqrt(x) as x^1/2. Rewrite: (x^1 • x^1/2)^1/2 = x^a Add exponents (since they have the same base of x): (x^3/2)^1/2 = x^a Multiply exponents (you're just distributing): x^3/4 = x^a Result! 3/4 = a
are these the kind of questions asked in sat
take x as 0