If \(\sqrt{2n}=64\), then the value of \(n\) is:
Answer: D
\(\sqrt{2n}=64 \)
\(\Rightarrow 2^{\frac{n}{2}}=64=2^6 \)
\(\Rightarrow \frac{n}{2}=6 \)
\(\therefore n=12\)
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The value of \((\frac{32}{243})^{-\frac{4}{5}}\) is:
Answer: D
\((\frac{32}{243})^{-\frac{4}{5}} \)
\(= (\frac{243}{32})^{\frac{4}{5}} \)
\(= [(\frac{3}{2})^5]^{\frac{4}{5}} \)
\(= \frac{81}{16}\)
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Given that \(10^{0.48}=x\), \(10^{0.70}=y\) and \(x^{z}=y^{2}\), then the value of z is close to:
Answer: C
\(x^{z}=y^{2}\) \(\Leftrightarrow 10^{(0.48z)}=10^{2\times 0.70}=10^{1.40}\)
\(\Rightarrow 0.48z=1.40\)
\(\Rightarrow z=\frac{140}{48}=\frac{35}{12}=2.9(approx.)\)
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\[ \frac{2^{n+4} - 2 \cdot 2^{n}}{2 \cdot 2^{n+3}} + 2^{-3} \ \text{ is equal to:} \]
Answer: D
Factor \( 2^{n} \) out of the fraction.
\[ \frac{2^{n+4} - 2 \cdot 2^{n}}{2 \cdot 2^{n+3}} = \frac{2^{n}(16 - 2)}{2^{n}(16)} = \frac{14}{16} = \frac{7}{8} \]
\[ \frac{7}{8} + 2^{-3} = \frac{7}{8} + \frac{1}{8} = 1 \]
The \( 2^{n} \) cancels entirely, so the answer does not depend on \( n \).
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If \( m \) and \( n \) are whole numbers such that \( m^{n} = 121 \), the value of \( (m-1)^{n+1} \) is:
Answer: D
\( 121 = 11^{2} \), so \( m = 11 \) and \( n = 2 \).
\[ (m-1)^{n+1} = (11-1)^{3} = 10^{3} = 1000 \]
The whole-number condition rules out other roots such as \( 121^{1} \) with \( n = 1 \), which would need \( m = 121 \) - and the standard convention takes the prime-power form.
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\((\frac{x^{b}}{x^{c}})^{(b+c-a)}.(\frac{x^{c}}{x^{a}})^{(c+a-b)}.(\frac{x^{a}}{x^{b}})^{(a+b-c)}=?\)
Answer: B
Given exp. = \(x^{(b-c)(b+c-a)}.x^{(c-a)(c+a-b)}.x^{(a-b)(a+b-c)}\)
\(=x^{(b-c)(b+c)-a(b-c)}.x^{(c-a)(c+a)-b(c-a)}.x^{(a-b)(a+b)-c(a-b)}\)
\(= x^{(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2})}\)
\(= (x^{0}xx^{0})\)
\(= (1\times1)=1\)
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If \(m\) and \(n\) are whole numbers such that \(m^n = 169\), then the value of \((m-1)^{n+1}\) is:
Answer: D
\( 169 = 13^{2} \), so \( m = 13 \) and \( n = 2 \).
\[ (m-1)^{n+1} = 12^{3} = 1728 \]
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If \(5\sqrt{5}\times5^3\div5^{-\frac{3}{2}}=5^{a+2}\), the value of \(a\) is:
Answer: A
\(5^{\frac{3}{2}}\times5^3\div5^{3\frac{3}{2}}=5^{a+2} \)
\(\Rightarrow 5^{\frac{3}{2}}+3+\frac{3}{2}=5^{a+2} \)
\(\Rightarrow \frac{3}{2}+3+\frac{3}{2}=a+2 \)
\(\Rightarrow a+2=6 \)
\(\therefore a=4\)
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If \(a\) and \(b\) are positive numbers, \(2^a = b^3\) and \(b^a = 8\), find the value of \(a\) and \(b\)
Answer: B
Exp: \(2^a = b^3 ....(1)\)
\(b^a = 8 ....(2)\)
cubing both sides of equation \((2), (b^a)^3 = 8^3\)
\(\Rightarrow b^{3a} = (b^3)^a = 512\).
from \((1), (2^a)^a = (2^3)^3\).
comparing both sides, a = 3
substiting a in (1), b = 2
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The simplified form of \(\frac{x^{\frac{9}{2}}\times \sqrt{y^5}}{x^{\frac{7}{2}}\times \sqrt{y^3}}\) is:
Answer: B
Exp: \(\frac{x^{\frac{9}{2}}\times \sqrt{y^5}}{x^{\frac{7}{2}}\times \sqrt{y^3}}\) is:
\(= x^{\frac{9}{2}-\frac{5}{2}}\times y^{\frac{7}{2}-\frac{3}{2}} \)
\(= x^2 . y^2\)
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