Alg9.4.3

If a not equal to 1 and m,k are positive real numbers, then $\log_a(m^k)=k\cdot\log_a(m)\,$

Let

$\log_a(m)=x\,$

$m=a^x\,$

$m^k=(a^x)^{k}\,$

$m^k=a^{kx}\,$

Taking log on both sides with base a,we get

$\log_a(m^k)=kx\,$

Substituting the value of x in the above,we get

$\log_a(m^k)=k\cdot\log_a(m)\,$