SOLUTION: what is x if: logx(16)=-2 I have x^-2=16 x^-2^-1=16^-1 x^2=1/16 2 squareroot 1/16

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Question 240520: what is x if: logx(16)=-2
I have x^-2=16
x^-2^-1=16^-1
x^2=1/16
2 squareroot 1/16
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28x%2C+%2816%29%29+=+-2
All of this...
I have x%5E-2=16
x%5E-2%5E-1=16%5E-1
x%5E2=1%2F16
x+=+sqrt%281%2F16%29 (Rejecting -sqrt(1/16) because x is the base of a logarithm and we don't have logarithms with negative bases.)
... is great. But you should simplify your answer:
x+=+sqrt%281%2F16%29+=+sqrt%281%29%2Fsqrt%2816%29+=+1%2F4