The value of $$\left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right)$$ is = ?
A. $$\frac{3}{2}$$
B. $$\frac{4}{9}$$
C. $$\frac{{16}}{{81}}$$
D. $$\frac{{32}}{{243}}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right) \cr & = \frac{{{9^2} \times {{\left( {9 \times 2} \right)}^4}}}{{{3^{16}}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^2} \times {{\left( {{3^2}} \right)}^4} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^4} \times {3^8} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{\left( {4 + 8} \right)}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{12}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{2^4}}}{{{3^{\left( {16 - 12} \right)}}}} \cr & = \frac{{{2^4}}}{{{3^4}}} \cr & = \frac{{16}}{{81}} \cr} $$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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