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

Question Number 65934 by gunawan last updated on 06/Aug/19
$$\:\mathrm{log}_{\mathrm{7}} \:\mathrm{2}={a} \\$$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}={b} \\$$$$\:\mathrm{log}_{\mathrm{6}} \:\mathrm{98}=… \\$$
Answered by meme last updated on 06/Aug/19
$$\:\:\:\:\:\:\:\:\:\:\:\:{answers} \\$$$$\:\:\:\:\:\:\:\:{log}_{\mathrm{7}} \mathrm{2}\:=\:\frac{{ln}\mathrm{2}}{{ln}\mathrm{7}}\:=\:\mathrm{0}.\mathrm{35}\:=\:{a} \\$$$$\:\:\:\:\:\:\:\:{log}_{\mathrm{2}} \mathrm{3}\:=\:\frac{{ln}\mathrm{3}}{{ln}\mathrm{2}}\:=\:\mathrm{1}.\mathrm{58}={b} \\$$$$\:\:\:\:\:\:\:\:{log}_{\mathrm{6}} \mathrm{98}=\frac{{ln}\mathrm{98}}{{ln}\mathrm{6}}=\:\mathrm{2}.\mathrm{55} \\$$
Answered by Kunal12588 last updated on 06/Aug/19
$${log}_{\mathrm{6}} \mathrm{98}={log}_{\mathrm{6}} \left(\mathrm{2}×\mathrm{7}^{\mathrm{2}} \right) \\$$$$={log}_{\mathrm{6}} \mathrm{2}+\mathrm{2}{log}_{\mathrm{6}} \mathrm{7} \\$$$$=\frac{{log}_{\mathrm{7}} \mathrm{2}}{{log}_{\mathrm{7}} \left(\mathrm{2}×\mathrm{3}\right)}+\frac{\mathrm{2}{log}_{\mathrm{2}} \mathrm{7}}{{log}_{\mathrm{2}} \mathrm{2}+{log}_{\mathrm{2}} \mathrm{3}} \\$$$$=\frac{{a}}{{a}+{log}_{\mathrm{7}} \mathrm{3}}+\frac{\mathrm{2}×\:\frac{{log}_{\mathrm{7}} \mathrm{7}}{{log}_{\mathrm{7}} \mathrm{2}}}{\mathrm{1}+{b}} \\$$$$=\frac{{a}}{{a}+\frac{{log}_{\mathrm{2}} \mathrm{3}}{{log}_{\mathrm{2}} \mathrm{7}}}+\frac{\mathrm{2}}{{a}\left(\mathrm{1}+{b}\right)} \\$$$$\Rightarrow\frac{{a}}{{a}+{ab}}+\frac{\mathrm{2}}{{a}+{ab}} \\$$$$\Rightarrow\frac{{a}+\mathrm{2}}{{a}+{ab}} \\$$
Answered by Kunal12588 last updated on 06/Aug/19
$${log}_{\mathrm{7}} \:\mathrm{2}=\frac{{ln}\:\mathrm{2}}{{ln}\mathrm{7}}={a} \\$$$${log}_{\mathrm{2}} \:\mathrm{3}=\frac{{ln}\:\mathrm{3}}{{ln}\:\mathrm{2}}={b} \\$$$${log}_{\mathrm{6}} \mathrm{98}=\frac{{ln}\:\mathrm{2}\:+\mathrm{2}\:{ln}\:\mathrm{7}\:}{{ln}\:\mathrm{2}\:+\:{ln}\:\mathrm{3}} \\$$$$=\frac{\frac{{ln}\:\mathrm{2}}{{ln}\:\mathrm{2}}+\frac{\mathrm{2}\:{ln}\:\mathrm{7}}{{ln}\:\mathrm{2}}}{\frac{{ln}\:\mathrm{2}}{{ln}\:\mathrm{2}}+\frac{{ln}\:\mathrm{3}}{{ln}\:\mathrm{2}}} \\$$$$=\frac{\mathrm{1}+\frac{\mathrm{2}}{{a}}}{\mathrm{1}+{b}} \\$$$$=\frac{{a}+\mathrm{2}}{{a}+{ab}} \\$$
Commented by gunawan last updated on 06/Aug/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\$$