$$\frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} = ?$$
A. 1
B. 2
C. 9
D. 3n
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {\text{Given}}\,{\text{Expression}} \cr & = \frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{{\left( {{3^2}} \right)}^n} \times {3^{n - 1}}}} \cr & = \frac{{\left( {{3^{5 \times \left( {n/5} \right)}} \times {3^{2n + 1}}} \right)}}{{\left( {{3^{2n}} \times {3^{n - 1}}} \right)}} \cr & = \frac{{{3^n} \times {3^{2n + 1}}}}{{{3^{2n}} \times {3^{n - 1}}}} \cr & = \frac{{{3^{\left( {n + 2n + 1} \right)}}}}{{{3^{\left( {2n + n - 1} \right)}}}} \cr & = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & = {3^{\left( {3n + 1 - 3n + 1} \right)}} \cr & = {3^2} \cr & = 9 \cr} $$Related Questions on Surds and Indices
A. $$\frac{1}{2}$$
B. 1
C. 2
D. $$\frac{7}{2}$$
Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7
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