Question Number 160939 by mathlove last updated on 09/Dec/21
$${log}\underset{{ab}} {{x}}=?\:\:\:\:\:\:{if}\:\:{log}\underset{{a}} {{x}}=\mathrm{30}\:\:{andlog}\underset{{b}} {{x}}=\mathrm{70} \\ $$
Commented by blackmamba last updated on 09/Dec/21
$$\:\begin{cases}{{a}={x}^{\frac{\mathrm{1}}{\mathrm{30}}} }\\{{b}={x}^{\frac{\mathrm{1}}{\mathrm{70}}} }\end{cases}\:\Rightarrow\:\mathrm{log}\:_{{ab}} \left({x}\right)=\:\mathrm{log}\:_{{x}^{\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{70}}} } \left({x}\right)=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{70}}}\:=\:\mathrm{21} \\ $$
Answered by qaz last updated on 09/Dec/21
$$\mathrm{log}_{\mathrm{ab}} \mathrm{x}=\frac{\mathrm{lnx}}{\mathrm{ln}\left(\mathrm{ab}\right)}=\frac{\mathrm{lnx}}{\mathrm{lna}+\mathrm{lnb}}=\frac{\mathrm{1}}{\frac{\mathrm{lna}}{\mathrm{lnx}}+\frac{\mathrm{lnb}}{\mathrm{lnx}}}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{a}} \mathrm{x}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{b}} \mathrm{x}}}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{70}}}=\mathrm{21} \\ $$
Answered by cortano last updated on 09/Dec/21
$$\:\begin{cases}{\mathrm{log}\:_{\mathrm{a}} \left(\mathrm{x}\right)=\mathrm{30}}\\{\mathrm{log}\:_{\mathrm{b}} \left(\mathrm{x}\right)=\mathrm{70}}\end{cases} \\ $$$$\:\:\:\mathrm{log}\:_{\mathrm{ab}} \left(\mathrm{x}\right)=\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{ab}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{a}\right)+\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{b}\right)} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{70}}}\:=\:\frac{\mathrm{30}×\mathrm{70}}{\mathrm{100}}\:=\:\mathrm{21} \\ $$