Question 829896
{{{log ((1/(10^-3)))+log (10^4)}}}
Since {{{(1/(10^-3)) = (10^3)}}}, the general rule is {{{x^(-1)=(1/x)}}}
{{{log ((10^3))+log ((10^4))}}}
{{{log ((10^7))}}}
Since {{{log ((10^7))}}} means the exponent on 10 that produces {{{10^7}}},
{{{log ((10^7))}}}=7