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CN-LEAGUE 1998 Problem 1

    If a>1,b>1,lg⁡(a+b)=lg⁡a+lg⁡ba>1,b>1,\lg(a+b)=\lg a+\lg b , then the value of lg⁡(a−1)+lg⁡(b−1)\lg(a-1)+\lg(b-1) is (A)lg⁡2(B)1(C)0(D)\text{(A)}\lg2 \text{(B)}1 \text{(C)}0 \text{(D)} not sure

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