We have, #10; #10; #160; 1=2^( #10;2n+m ) #10; #10; #160; #10;2^0=2^( 2n+m ) ∴ #10; ( some )^0=1 #10; #10;Comparing t ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

(a-b)^2 Expansion and Visualisation

Solve 1 = 2...

Question

Solve $1 = 2^{(2 n + m)}$

Open in App

Solution

We have,

$1 = 2^{(2 n + m)}$

$2^{0} = 2^{(2 n + m)} ∴ {(s o m e)}^{0} = 1$

Comparing that and we get,

$2 n + m = 0$

$\Rightarrow 2 n = - m$

$\Rightarrow - m = 2 n$

$\Rightarrow \frac{m}{n} = \frac{- 2}{1}$

$\Rightarrow m : n = - 2 : 1$

Hence, this is the answer.

Suggest Corrections

Similar questions

L i m i t n \to \infty \frac{5^{n + 1} + 3^{n} - 2^{2 n}}{5^{n} + 2^{n} + 3^{2 n + 3}} =

{c o t}^{- 1} (2^{2} + \frac{1}{2}) + {c o t}^{- 1} (2^{3} + \frac{1}{2^{2}}) + {c o t}^{- 1} (2^{4} + \frac{1}{2^{3}}) + . . . . . \infty

{t a n}^{- 1} \frac{1}{3} + {t a n}^{- 1} \frac{2}{9} + . . . + {t a n}^{- 1} \frac{2^{n - 1}}{1 + 2^{2 n - 1}}

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

\sum \sum 0 \leq r < s \leq n (r + s) C_{r} C_{s}

\sum_{r = 1}^{n} r .^{2 n} C_{r}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

Join BYJU'S Learning Program