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Answer and get the brainliest answer reward |
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Answer» Easy 2^a=3^b=6^(-c)=k 2^a=k k^(1/a)=2 similarly k^(1/b)=3 & k^(-1/c)=6 and since 2×3=6 [k^(1/a)]×[k^(1/b)]=k^(-1/c) k^[(1/a)+(1/b)]=k^(-1/c) since BASES are same thats why powers will be equal therefore, (1/a)+(1/b)= -1/c (1/a)+(1/b)+(1/c)=0 Mark me as the BRAINLIEST |
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