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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If \(2^{(x-y)} = 8\) and \(2^{(x+y)} = 32\), then \(x\) is equal to:
Answer: B
\(2^{(x-y)} = 8 = 2^3 \)
\(\Rightarrow x-y = 3 ....(1) \)
\(2^{(x+y)} = 32 = 2^5 \)
\(\Rightarrow x+y = 5 ....(2)\)
On solving (1) & (2), we get \(x = 4.\)
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\((0.04)^{-1.5}=?\)
Answer: B
\((0.04)^{-1.5}=(\frac{4}{100})^{-1.5}\)
\(=(\frac{1}{25})^{-(3/2)}\)
\(=(25)^{(3/2)}\)
\(=(5)^{2\times(3/2)}\)
\(=5^{3}\)
=125
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If \((\frac{a}{b})^{x-1}=(\frac{b}{a})^{x-3}\), then the value of \(x\) is:
Answer: C
Given \((\frac{a}{b})^{x-1}=(\frac{b}{a})^{x-3}\)
\(\Rightarrow (\frac{a}{b})^{x-1}=(\frac{a}{b})^{-(x-3)}=(\frac{a}{b})^{(3-x)} \)
\(\Rightarrow x-1=3-x \)
\(\Rightarrow 2x=4 \)
\(\Rightarrow x=2\)
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The value of \([(10)^{150}\div (10)^{146}]\)
Answer: B
\((10)^{150}\div (10)^{146}=\frac{10^{150}}{10^{146}}\)
\(=10^{150-146}\)
\(=10^{4}\)
\(=10000\)
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If \(3^x - 3^{x-1} = 18\), then the value of \(x^x\) is:
Answer: A
\(3^x - 3^{x-1} = 18 \)
\(\Rightarrow 3^{x-1} \times (3-1) = 18 \)
\(\Rightarrow 3^{x-1} = 9 = 3^2 \)
\(\Rightarrow x-1 = 2 \)
\(\Rightarrow x = 3\).
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if \(3^{(x-y)}=27\) and \(3^{(x+y)}=243\), then \(x\) is equal to:
Answer: B
\(3^{x-y} =27 = 3^{3} \Leftrightarrow x-y=3\) ...(i)
\(3^{x+y}=243=3^{5} \Leftrightarrow x+y=5\) ...(ii)
On solving (i) and (ii), we get \(x=4\).
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Solve for \(((3^{y})^{\sqrt{}2})^{2} = 729\).
Answer: C
Exp: \(((3^{y})^{\sqrt{}2})^{2} = 729\)
\((3^y)^2 = 3^4 (\sqrt2^2 = (2^{\frac{1}{2}}) = 2)\)
equating powers of 2 on both sides,
\(y^2 = 4\)
\(\Rightarrow y = \pm2\)
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\(\sqrt{[200\sqrt{[200\sqrt{[200....... \infty ]}]}]} = ?\)
Answer: A
Exp: Let \(\sqrt{[200\sqrt{[200\sqrt{[200....... \infty ]}]}]} = x\).
Hence \(\sqrt{200x} = x\)
Squaring both sides \(200x = x^2\)
\(\Rightarrow x(x-200) = 0\)
\(\Rightarrow x = 0 \) or \(x-200 = 0\) i.e. \(x = 200\)
As \(x\) cannot be 0, \(x = 200\)
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