જો ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ તો $m =$
Easy
Download our app for free and get started
d
(d) \({2^{m(n + 1) + 2n + n}} = {2^{(m + 1)n + 2m}}\)
(d) \({2^{m(n + 1) + 2n + n}} = {2^{(m + 1)n + 2m}}\)
\( \Rightarrow \)\(mn + m + 3n = mn + 2m + n\)\( \Rightarrow \)\(m = 2n\).
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $View Solution
- 2જો $x = {\log _b}a,\,\,y = {\log _c}b,\,\,\,z = {\log _a}c$ તો $xyz = . . . .$View Solution
- 3જો ${\log _k}x.\,{\log _5}k = {\log _x}5,k \ne 1,k > 0$ તો $x = . . . .$View Solution
- 4${{\sqrt {6 + 2\sqrt 3 + 2\sqrt 2 + 2\sqrt 6 } - 1} \over {\sqrt {5 + 2\sqrt 6 } }}$View Solution
- 5જો ${{ax - 1} \over {(1 - x + {x^2})\,(2 + x)}} = {x \over {1 - x + {x^2}}} - {1 \over {2 + x}}$, તો $a = $View Solution
- 6${\log _2}7$ એ . . . . થાય.View Solution
- 7જો ${{{x^3} - 6{x^2} + 10x - 2} \over {{x^2} - 5x + 6}} = f(x) + {A \over {(x - 2)}} + {B \over {(x - 3)}}$, તો $f(x) = $View Solution
- 8${{{x^2} + 13x + 15} \over {(2x + 3)\,{{(x + 3)}^2}}} = $View Solution
- 9જો ${{1 - \cos x} \over {\cos x(1 + \cos x)}} = {{\sin \alpha } \over {\cos x}} - {2 \over {1 + \cos x}}$, તો $\alpha = $View Solution
- 10સમીકરણ ${\log _7}{\log _5}$ $(\sqrt {{x^2} + 5 + x} ) = 0$ નો ઉકેલ મેળવો.View Solution