SOLUTION: evaluate the exponential equation for three positive values of x, three negative values of x and at x=0 y=3^(x-4)

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Question 269444: evaluate the exponential equation for three positive values of x, three negative values of x and at x=0
y=3^(x-4)
Found 2 solutions by drk, JBarnum:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
we can use a table as follows:
x . . . -3 . . . . . . . . . . .-2 . . . . . . . . . -1 . . . . . . . .0 . . . . . . . . .1 . . . . . . . 2 . . . . . . .3
y . . . (0.000457) . . . . (0.00137) . . . .(0.00412) . . .(0.01234) . . .(0.037) . . .(0.11) . . . . (0.333)

Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
evaluate the exponential equation for three positive values of x, three negative values of x and at x=0
y=3%5E%28x-4%29
this is fun u get to pick ur x's
id pick (x=4,5,6)(x=-.1,-.01,-.001)(x=0) now just plug them in and solve for y
y=3%5E%284-4%29
y=3%5E%280%29
y=1
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y=3%5E%285-4%29
y=3%5E%281%29
y=3
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y=3%5E%286-4%29
y=3%5E%282%29
y=9
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so for (x=4,5,6) we have (y=1,3,9)
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y=3%5E%28-.1-4%29
y=3%5E%28-4.1%29
y=0.01106121555
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y=3%5E%28-.01-4%29
y=3%5E%28-4.01%29
y=0.01221079017
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y=3%5E%28-0.001-4%29
y=3%5E%28-4.001%29
y=0.01233212334
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so for (x=-.1,-.01,-.001) we have (y=0.0111,0.0122,0.0123)
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y=3%5E%280-4%29
y=3%5E%28-4%29
y=0.012345679+repeating
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for x=0, y=0.012345679