What is expand n(log_{10}(1-p)=-log_{10}(3)) ?

The solution to expand n(log_{10}(1-p)=-log_{10}(3)) is nlog_{10}(1-p)=-log_{10}(3)n

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expand n(log_{10}(1-p)=-log_{10}(3))

expand

n (lo g_{10} (1 - p) = - lo g_{10} (3))

n lo g_{10} (1 - p) = - lo g_{10} (3) n

Solution steps

n (lo g_{10} (1 - p) = - lo g_{10} (3))

= n (= lo g_{10} (1 - p) - lo g_{10} (3))

Apply the distributive law:

a (b - c) = ab - a c

a = n, b = lo g_{10} (1 - p) =, c = lo g_{10} (3)

n lo g_{10} (1 - p) = - n lo g_{10} (3)

n lo g_{10} (1 - p) = - lo g_{10} (3) n

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What is expand n(log_{10}(1-p)=-log_{10}(3)) ?
The solution to expand n(log_{10}(1-p)=-log_{10}(3)) is nlog_{10}(1-p)=-log_{10}(3)n