Question Number 170427 by mathlove last updated on 23/May/22
$$\mathrm{14}^{{b}} =\mathrm{7}^{{b}−{a}} \\ $$$$\mathrm{49}^{\frac{{a}}{{b}}} =? \\ $$
Answered by thfchristopher last updated on 23/May/22
![14^b =7^(b−a) ⇒(2^b )(7^b )=7^(b−a) ⇒log_7 [(2^b )(7^b )]=log_7 7^(b−a) ⇒blog_7 2+blog_7 7=blog_7 7−alog_7 7 ⇒blog_7 2=−a ⇒(a/b)=−log_7 2 ∴ 49^(a/b) =(7^2 )^(−log_7 2) =(1/7^(log_7 4) ) =(1/4)](Q170431.png)
$$\mathrm{14}^{{b}} =\mathrm{7}^{{b}−{a}} \\ $$$$\Rightarrow\left(\mathrm{2}^{{b}} \right)\left(\mathrm{7}^{{b}} \right)=\mathrm{7}^{{b}−{a}} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{7}} \left[\left(\mathrm{2}^{{b}} \right)\left(\mathrm{7}^{{b}} \right)\right]=\mathrm{log}_{\mathrm{7}} \mathrm{7}^{{b}−{a}} \\ $$$$\Rightarrow{b}\mathrm{log}_{\mathrm{7}} \mathrm{2}+{b}\mathrm{log}_{\mathrm{7}} \mathrm{7}={b}\mathrm{log}_{\mathrm{7}} \mathrm{7}−{a}\mathrm{log}_{\mathrm{7}} \mathrm{7} \\ $$$$\Rightarrow{b}\mathrm{log}_{\mathrm{7}} \mathrm{2}=−{a} \\ $$$$\Rightarrow\frac{{a}}{{b}}=−\mathrm{log}_{\mathrm{7}} \mathrm{2} \\ $$$$\therefore\:\mathrm{49}^{\frac{{a}}{{b}}} \\ $$$$=\left(\mathrm{7}^{\mathrm{2}} \right)^{−\mathrm{log}_{\mathrm{7}} \mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{log}_{\mathrm{7}} \mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}} \\ $$