# Answer to Question #12637 in Calculus for Bryan Grasso

Question #12637

Log I = Log 10^-16 + 9....Im having trouble solving for I. I understand Log 10 to any power is that number (in this case, -16). But I still am having trouble solving for I

Expert's answer

Log I = Log 10^-16 + 9

As you've mentioned absolutely correctly, Log 10^-16=-16*Log 10=-16

Therefore we have Log I = Log 10^-16 + 9=-16+9=-8

So due to definition of logarithm I=10^-8

As you've mentioned absolutely correctly, Log 10^-16=-16*Log 10=-16

Therefore we have Log I = Log 10^-16 + 9=-16+9=-8

So due to definition of logarithm I=10^-8

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