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