Question 1128806
After 14 years, from what?


{{{y=10*b^x}}}

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{{{10*b^9=5}}}

{{{b^9=1/2}}}

{{{log((b^9))=log((1/2))}}}

{{{9*log((b))=log((1/2))}}}

{{{log((b))=(1/9)*log((1/2))}}}

{{{log((b))=log(((0.5)^(1/9)))}}}

{{{b=(0.5)^(1/9)}}}


{{{b=0.9259}}}


{{{highlight(y=10*(0.9259)^x)}}}--------the model to use.




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this is not the only exponential model possible.
notice that the given data is for half-life.