Given, 5log10(25)=(2.5)log105⇒5x−1=(5)log10(2.5)[∵alog(b)=blog(a)] On comparing LHS and RHS, we get x−1=log10(2.5)x−1=log10(1025)x−1=log1025−log1010[∵log(nm)=logm−logn]x−1=log1052−1[∵logaa=1]x=log1052x=2log105[∵logam=mloga] Hence, the value of x is 2log105.