Question

simplify [243^(x/5) * 3^2x]/[9^x*3^x]

Answer 1

We have been given the term to solve,

\frac{\frac{243^{n}}{5.3^{2n+1}}}{9^{n}\times 3^{n-1}}

9

n

×3

n−1

5.3

2n+1

243

n

So, on solving the equation using the BODMAS rule, we get,

\frac{243^{n}}{5.3^{2n+1}}\times \frac{1}{9^{n}\times 3^{n-1}}=\frac{3^{5}}{15.3^{2n}\times 3^{2n}\times 3^{n-1}}=\frac{3^{5}}{5(3^{2n+1}\times 3^{2n}\times 3^{n-1})}

5.3

2n+1

243

n

×

9

n

×3

n−1

1

=

15.3

2n

×3

2n

×3

n−1

3

5

=

5(3

2n+1

×3

2n

×3

n−1

)

3

5

Now, we also know that,

xᵃ × xᵇ = xᵃ⁺ᵇ

Therefore, on using the same identity in the equation, we get,

\begin{gathered}\frac{3^{5}}{5(3^{2n+1}\times 3^{2n}\times 3^{n-1})}=\frac{3^{5}}{5(3^{5n})}\\\frac{3^{5}}{5(3^{5n})}=\frac{3^{5-5n}}{5}\end{gathered}

5(3

2n+1

×3

2n

×3

n−1

)

3

5

=

5(3

5n

)

3

5

5(3

5n

)

3

5

=

5

3

5−5n

Because,

\frac{a^{m}}{a^{n}}=a^{m-n}

a

n

a

m

=a

m−n

Therefore, the final solution of the equation on simplifying is given by,

\frac{\frac{243^{n}}{5.3^{2n+1}}}{9^{n}\times 3^{n-1}}=\frac{3^{5-5n}}{5}

9

n

×3

n−1

5.3

2n+1

243

n

=

5

3

5−5n