Question

how to find log (1/10^-2.3)​

Answer 1

$log(1/10^-^2^.^3)\ is\ in\ the\ form:\\log(1*10^2^.^3)\\\boxed{log(a.b)=log(a)+log(b)}\\\boxed{log(a^b)=b*log(a)}\\Now,\\log(1*10^2^.^3)=log(1)+log(10^2^.^3)\\=log(1)+2.3*log(10)\\=0+2.3*1\\\bold{[As\ log(1)=0\ and\ log(10)=1]}\\=2.3$

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