Question Number 84000 by byaw last updated on 08/Mar/20
Commented by Kunal12588 last updated on 08/Mar/20
$${log}\:\sqrt[{\mathrm{3}}]{\mathrm{0}.\mathrm{0005957}}\:{should}\:{be}\:−{ve}\:? \\ $$
Answered by MJS last updated on 08/Mar/20
![log_b 5.957 =.7750 ⇔ 5.957=b^(.7750) (((5.957)/(10000)))^(1/3) =b^x x=((ln (((5.957)/(10000)))^(1/3) )/(ln b))=(((1/3)(ln 5.957 −4ln 10))/(ln b))= =(1/3)(((ln 5.957)/(ln b))−((4ln 10)/(ln b)))=(1/3)(log_b 5.957 −4log_b 10)= =(1/3)(.7750−4log_b 10)= [b^(.7750) =5.957 ⇒ b≈10] =−1.075](https://www.tinkutara.com/question/Q84044.png)
$$\mathrm{log}_{{b}} \:\mathrm{5}.\mathrm{957}\:=.\mathrm{7750} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{5}.\mathrm{957}={b}^{.\mathrm{7750}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{5}.\mathrm{957}}{\mathrm{10000}}}={b}^{{x}} \\ $$$${x}=\frac{\mathrm{ln}\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{5}.\mathrm{957}}{\mathrm{10000}}}}{\mathrm{ln}\:{b}}=\frac{\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{ln}\:\mathrm{5}.\mathrm{957}\:−\mathrm{4ln}\:\mathrm{10}\right)}{\mathrm{ln}\:{b}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{ln}\:\mathrm{5}.\mathrm{957}}{\mathrm{ln}\:{b}}−\frac{\mathrm{4ln}\:\mathrm{10}}{\mathrm{ln}\:{b}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}_{{b}} \:\mathrm{5}.\mathrm{957}\:−\mathrm{4log}_{{b}} \:\mathrm{10}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(.\mathrm{7750}−\mathrm{4log}_{{b}} \:\mathrm{10}\right)= \\ $$$$\:\:\:\:\:\left[{b}^{.\mathrm{7750}} =\mathrm{5}.\mathrm{957}\:\Rightarrow\:{b}\approx\mathrm{10}\right] \\ $$$$=−\mathrm{1}.\mathrm{075} \\ $$