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(Solved): \( m^{i}: m^{j}=m^{i+j} \) \( \left(m^{i}\right)^{j}=m^{(i \cdot j)} \) ) \( \left(m^{i}\right) \cd ...



\( m^{i}: m^{j}=m^{i+j} \) \( \left(m^{i}\right)^{j}=m^{(i \cdot j)} \) ) \( \left(m^{i}\right) \cdot(n i)=(m \cdot n)^{i} \)
\( m^{i}: m^{j}=m^{i+j} \) \( \left(m^{i}\right)^{j}=m^{(i \cdot j)} \) ) \( \left(m^{i}\right) \cdot(n i)=(m \cdot n)^{i} \)


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(i) mi.mj=mi+j Prove Taking log of Both We get log?(mi+j)=log?(mi.mj)(i+j)log?m=log?(mi)+log?(mj)ilog?m+jlog?m=ilog?m+jlog?m Proved
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