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Question Number 65970 by gunawan last updated on 07/Aug/19

log_5 (√(27))×log_9 125+log_(16) 12=... a. ((61)/(36)) b. (9/4) c. ((61)/(20)) d. ((41)/(12)) e. (7/2)

$$\mathrm{log}_{\mathrm{5}} \sqrt{\mathrm{27}}×\mathrm{log}_{\mathrm{9}} \mathrm{125}+\mathrm{log}_{\mathrm{16}} \mathrm{12}=... \\ $$$${a}.\:\frac{\mathrm{61}}{\mathrm{36}} \\ $$$${b}.\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${c}.\:\frac{\mathrm{61}}{\mathrm{20}} \\ $$$${d}.\:\frac{\mathrm{41}}{\mathrm{12}} \\ $$$${e}.\:\frac{\mathrm{7}}{\mathrm{2}} \\ $$

Answered by MJS last updated on 07/Aug/19

log_b a =((ln a)/(ln b)) ((ln (√(27)))/(ln 5))×((ln 125)/(ln 9))+((ln 12)/(ln 16))=((ln 3^(3/2) )/(ln 5))×((ln 5^3 )/(ln 3^2 ))+((ln 2^2 3)/(ln 2^4 ))= =(((3/2)ln 3)/(ln 5))×((3ln 5)/(2ln 3))+((2ln 2 +ln 3)/(4ln 2))=(9/4)+(1/2)+((ln 3)/(4ln 2))= =((11)/4)+((ln 3)/(4ln 2)) please check given equation

$$\mathrm{log}_{{b}} \:{a}\:=\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{b}} \\ $$$$\frac{\mathrm{ln}\:\sqrt{\mathrm{27}}}{\mathrm{ln}\:\mathrm{5}}×\frac{\mathrm{ln}\:\mathrm{125}}{\mathrm{ln}\:\mathrm{9}}+\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{16}}=\frac{\mathrm{ln}\:\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{ln}\:\mathrm{5}}×\frac{\mathrm{ln}\:\mathrm{5}^{\mathrm{3}} }{\mathrm{ln}\:\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{ln}\:\mathrm{2}^{\mathrm{2}} \mathrm{3}}{\mathrm{ln}\:\mathrm{2}^{\mathrm{4}} }= \\ $$$$=\frac{\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{5}}×\frac{\mathrm{3ln}\:\mathrm{5}}{\mathrm{2ln}\:\mathrm{3}}+\frac{\mathrm{2ln}\:\mathrm{2}\:+\mathrm{ln}\:\mathrm{3}}{\mathrm{4ln}\:\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4ln}\:\mathrm{2}}= \\ $$$$=\frac{\mathrm{11}}{\mathrm{4}}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4ln}\:\mathrm{2}} \\ $$$$\mathrm{please}\:\mathrm{check}\:\mathrm{given}\:\mathrm{equation} \\ $$

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