log-7-2-a-log-2-3-b-log-6-98-

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Question Number 65934 by gunawan last updated on 06/Aug/19

$${\log }_{7}2=a$$$${\log }_{2}3=b$$$${\log }_{6}98=\dots$$

Answered by meme last updated on 06/Aug/19

$$answers$$$${log}_{7}2=\frac{ln2}{ln7}=0.35=a$$$${log}_{2}3=\frac{ln3}{ln2}=1.58=b$$$${log}_{6}98=\frac{ln98}{ln6}=2.55$$

Answered by Kunal12588 last updated on 06/Aug/19

$${log}_{6}98={log}_6\left(2\times 7^2\right)$$$$={log}_{6}2+2{log}_{6}7$$$$=\frac{{log}_{7}2}{{log}_7\left(2\times 3\right)}+\frac{2{log}_{2}7}{{log}_{2}2+{log}_{2}3}$$$$=\frac{a}{a+{log}_{7}3}+\frac{2\times \frac{{log}_{7}7}{{log}_{7}2}}{1+b}$$$$=\frac{a}{a+\frac{{log}_{2}3}{{log}_{2}7}}+\frac{2}{a(1+b)}$$$$\Rightarrow \frac{a}{a+ab}+\frac{2}{a+ab}$$$$\Rightarrow \frac{a+2}{a+ab}$$

Answered by Kunal12588 last updated on 06/Aug/19

$${log}_{7}2=\frac{ln2}{ln7}=a$$$${log}_{2}3=\frac{ln3}{ln2}=b$$$${log}_{6}98=\frac{ln2+2ln7}{ln2+ln3}$$$$=\frac{\frac{ln2}{ln2}+\frac{2ln7}{ln2}}{\frac{ln2}{ln2}+\frac{ln3}{ln2}}$$$$=\frac{1+\frac{2}{a}}{1+b}$$$$=\frac{a+2}{a+ab}$$

Commented by gunawan last updated on 06/Aug/19

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

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