find values of M and n if 4^2m = cube root of (16)^-6/n =( Square root of 8) ^2
Answers
Answered by
35
Solution:-
given by:-
![{4}^{2m} = \sqrt[3]{ {(16)}^{ - \frac{6}{n} } } = { (\sqrt{8} )}^{2} \\ \\ {2}^{4m} = {2}^{3} \\ 4m = 3 \\ (m = \frac{3}{4} ) \: ans \\ \\ \sqrt[3]{( {16)}^{ - \frac{6}{n} } } = {( \sqrt{8}) }^{2} \\ {16}^{ \frac{ - 6}{n} \times \frac{1}{3} } = {2}^{3} \\ {( {2}^{4} )}^{ \frac{ - 2}{n} } = {2}^{3} \\ {2}^{ - \frac{8}{n} } = {2}^{3} \\ \frac{ - 8}{n } = 3 \\ \frac{ - n}{8} = \frac{1}{3} \\ (n = - \frac{8}{3} ) \: ans \\ i \: hope \: its \: helpfull](https://tex.z-dn.net/?f=+%7B4%7D%5E%7B2m%7D++%3D++%5Csqrt%5B3%5D%7B+%7B%2816%29%7D%5E%7B+-++%5Cfrac%7B6%7D%7Bn%7D+%7D+%7D++%3D+++%7B+%28%5Csqrt%7B8%7D+%29%7D%5E%7B2%7D++++%5C%5C+%5C%5C++%7B2%7D%5E%7B4m%7D++%3D++%7B2%7D%5E%7B3%7D++%5C%5C+4m+%3D+3+%5C%5C+%28m+%3D++%5Cfrac%7B3%7D%7B4%7D+%29+%5C%3A+ans+%5C%5C++%5C%5C++%5Csqrt%5B3%5D%7B%28+%7B16%29%7D%5E%7B+-++%5Cfrac%7B6%7D%7Bn%7D+%7D+%7D++%3D++%7B%28+%5Csqrt%7B8%7D%29+%7D%5E%7B2%7D++%5C%5C++%7B16%7D%5E%7B+%5Cfrac%7B+-+6%7D%7Bn%7D++%5Ctimes++%5Cfrac%7B1%7D%7B3%7D+%7D++%3D++%7B2%7D%5E%7B3%7D++%5C%5C++%7B%28+%7B2%7D%5E%7B4%7D+%29%7D%5E%7B+%5Cfrac%7B+-+2%7D%7Bn%7D+%7D++%3D++%7B2%7D%5E%7B3%7D++%5C%5C++%7B2%7D%5E%7B+-++%5Cfrac%7B8%7D%7Bn%7D+%7D++%3D++%7B2%7D%5E%7B3%7D++%5C%5C++%5Cfrac%7B+-+8%7D%7Bn+%7D++%3D+3+%5C%5C++%5Cfrac%7B+-+n%7D%7B8%7D++%3D++%5Cfrac%7B1%7D%7B3%7D++%5C%5C+%28n+%3D++-++%5Cfrac%7B8%7D%7B3%7D+%29+%5C%3A+ans+%5C%5C+i+%5C%3A+hope+%5C%3A+its+%5C%3A+helpfull "{4}^{2m} = \sqrt[3]{ {(16)}^{ - \frac{6}{n} } } = { (\sqrt{8} )}^{2} \\ \\ {2}^{4m} = {2}^{3} \\ 4m = 3 \\ (m = \frac{3}{4} ) \: ans \\ \\ \sqrt[3]{( {16)}^{ - \frac{6}{n} } } = {( \sqrt{8}) }^{2} \\ {16}^{ \frac{ - 6}{n} \times \frac{1}{3} } = {2}^{3} \\ {( {2}^{4} )}^{ \frac{ - 2}{n} } = {2}^{3} \\ {2}^{ - \frac{8}{n} } = {2}^{3} \\ \frac{ - 8}{n } = 3 \\ \frac{ - n}{8} = \frac{1}{3} \\ (n = - \frac{8}{3} ) \: ans \\ i \: hope \: its \: helpfull")
given by:-
Similar questions