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Question Number 34301 by math1967 last updated on 03/May/18

If log_(12) 18=a ,find log_(24) 16 in term  of  a

$${If}\:{log}_{\mathrm{12}} \mathrm{18}={a}\:,{find}\:{log}_{\mathrm{24}} \mathrm{16}\:{in}\:{term} \\ $$$${of}\:\:{a} \\ $$

Answered by MJS last updated on 03/May/18

12^a =18  a=((ln 18)/(2ln 2+ln 3))  ((ln 18)/a)=2ln 2+ln 3  24^b =16  b=((ln 16)/(3ln 2+ln 3))  ((ln 16)/b)=3ln 2+ln 3  ((ln 16)/b)=((ln 18+aln 2)/a)  b=((aln 16)/(ln 18+aln 2))

$$\mathrm{12}^{{a}} =\mathrm{18} \\ $$$${a}=\frac{\mathrm{ln}\:\mathrm{18}}{\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{18}}{{a}}=\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3} \\ $$$$\mathrm{24}^{{b}} =\mathrm{16} \\ $$$${b}=\frac{\mathrm{ln}\:\mathrm{16}}{\mathrm{3ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{16}}{{b}}=\mathrm{3ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3} \\ $$$$\frac{\mathrm{ln}\:\mathrm{16}}{{b}}=\frac{\mathrm{ln}\:\mathrm{18}+{a}\mathrm{ln}\:\mathrm{2}}{{a}} \\ $$$${b}=\frac{{a}\mathrm{ln}\:\mathrm{16}}{\mathrm{ln}\:\mathrm{18}+{a}\mathrm{ln}\:\mathrm{2}} \\ $$

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