Show that {e^x, e^(x+2)} are linearly independent
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Let C1e^x + C2e^(x+2) = 0
for x = ln1: C1(1) + C2e^2 = 0, so C1 = -C2e^2
for x = 0: C1 + C2e^2 = 0
substituting gives: -C2e^2 + C2e^2 = 0
Does this suffice? Is there a way to get C1 = 0 and C2 = 0 explicitly?
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To show that the vectors {e^x, e^(x+2)} are linearly independent, we want to prove that the only solution to the equation C1e^x + C2e^(x+2) = 0 is when C1 = 0 and C2 = 0.
Let's solve the equation C1e^x + C2e^(x+2) = 0 step by step:
1. Substitute x = ln1 (which is 0) into the equation to get:
C1(1) + C2e^2 = 0
2. Simplify the equation to:
C1 + C2e^2 = 0
3. Now we substitute x = 0 into the equation to get:
C1 + C2e^2 = 0
4. Since the left side of the equation is the same as in step 2, we can equate them:
C1 + C2e^2 = -C2e^2
5. Now we can simplify the equation to:
C1 = -2C2e^2
6. If we substitute the value of C1 from step 5 into the equation of step 2, we get:
-2C2e^2 + C2e^2 = 0
7. Simplifying the equation gives us:
-C2e^2 = 0
From step 7, we can see that the only possible solution for C2 is 0. However, this does not immediately tell us the value of C1.
To determine the value of C1, we revisit the equation from step 5:
C1 = -2C2e^2
Substituting C2 = 0 yields:
C1 = 0
Therefore, we have found that the vectors {e^x, e^(x+2)} are linearly independent since the only solution to C1e^x + C2e^(x+2) = 0 is when C1 = 0 and C2 = 0.