SOLUTION: Could someone please tell me if I did this right. Evaluate the logarithmic equation for three values of x that are greater than 1, three values between 0 and 1, x to =0. y=log3

Question 211372: Could someone please tell me if I did this right.
Evaluate the logarithmic equation for three values of x that are greater than 1, three values between 0 and 1, x to =0.
y=log3.5^x
x=2 so y=0.55
x=3 so y=0.88
x=4 so y=1.11
x=0.2 so y= - 1.28
x=0.5 so y= - 0.55
x=0.7 so y= - 0.28
x=1 so y=0
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
y=log(3.5^x)
.
x=4,,,,,3.5^4=150,,,,,,log 150 = 2.176
x=3,,,,,3.5^3=42.9,,,,,log 42.9 =1.632
x=2,,,,,3.5^2=12.25,,,,log12.25 = 1.088
x=1,,,,,3.5^1 = 3.5,,,,log 3.5 = .544
x=.7,,,,3.5^.7=2.40,,,,log 2.40 = .3808
x=.5,,,,3.5^.5=1.87,,,,log 1.87=.2720
x=.2,,,,3.5^.2=1.28,,,,log 1.28 = .1088
.

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