Question

Solve the problem Guys..
Q23​

Answer 1

Answer:

$\frac{9^n\times 3^2\times(3^{(-n/2))^{-2}}-27^n}{3^{3m}\times 2^3}$

$\implies \frac{3^{2n}\times 3^2\times3^{n}-3^{3n}}{3^{3m}\times 2^3}$

$\implies \frac{3^{3n}(3^2-1)}{3^{3m}\times 2^3}$

$\implies \frac{3^{3n}(9-1)}{3^{3m}\times 2^3}$

$\implies \frac{3^{3n}(8)}{3^{3m}\times 2^3}$

$Given: \frac{3^{3n}(8)}{3^{3m}\times 2^3}=\frac{1}{27}\\\\\\\implies \frac{3^{3n}}{3^{3m}}=3^{-3}\\\\\\\implies 3^{3n-3m}=3^{-3}\\\\\\\implies 3n-3m=-3\\\\\\\implies m-n=1$

Explanation :

→ The laws of indices are used in the above sum .

→ $a^n\times a^m=a^{n+m}$

→ $\frac{a^n}{a^m}=a^{n-m}$

→ We will take commons and then cancel the denominator and numerator .

→ At last we can compare powers when base is equal .

→ $\mathsf{If\:a^n=a^m,then\:n=m}$

→ Note that a , n , m should not be 1 , or 0 .

Answer 2

ANSWER:------------

{33m×239n×32×(3(−n/2))−2−27n}

⟹{32n×32×3n−33n33m×23\implies \frac{3^{2n}\time)

(3^2\times3^{n}-3^{3n}}{3^{3m}\times 2^3}⟹33m×2332n×32×3n−33n)

(⟹33n(32−1)33m×23\implies \frac{3^{3n}(3^2-1)}{3^{3m}\times 2^3}⟹33m×2333n(32−1)

⟹(33n(9−1)33m×23\implies \frac{3^{3n}(9-1)}{3^{3m}\times 2^3}⟹33m×2333n(9−1)

⟹33n(8)33m×23\implies \frac{3^{3n}(8)}{3^{3m}\times 2^3}⟹33m×2333n(8)

{Given:33n(8)33m×23=127⟹33n33m=3−3⟹33n−3m=3−3⟹3n−3m=−3⟹m−n

(=1\begin{lgathered}Given: \frac{3^{3n}(8)}{3^{3m}\times)

2^3}=\frac{1}{27}\\\\\\\implies \frac{3^{

3n}}{3^{3m}}=3^{-3}\\\\\\\implies

3^{3n-3m}=3^{-3}\\\\\\\implies

3n-3m=-3\\\\\\\implies m-

n=1\end{lgathered}Given:33m×2333n(8)=271⟹33m33n

=3−3⟹33n−3m=3−3⟹3n−3m=−3⟹m−n

=1

hope it helps:----------

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