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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Simplify: \(\sqrt{5} \times \sqrt{20}\)
Answer: A
Correct answer: 10
Surds multiply under a single root sign.
\[\sqrt{5} \times \sqrt{20} = \sqrt{100} = 10\]
Answering 100 stops at the product inside the root without taking the square root.
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\((17)^{3.5}\times (17)^{?}=17^{8}\)
Answer: D
Let \((17)^{3.5}\times (17)^{x}=17^{8}\)
Then, \((17)^{3.5+x}=17^{8}\)
\(\therefore 3.5+x=8 \)
\(\Rightarrow x=(8-3.5) \)
\(\Rightarrow x=4.5\)
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If \(a^x = b^y = c^z\) and \(b^2 = ac\), then \(y\) equals:
Answer: D
Let \(a^x = b^y = c^z = k\).
Then, \(a = k^{\frac{1}{x}}, b = k^{\frac{1}{y}}, c = k^{\frac{1}{z}}\)
Therefore, \(b^2 = ac \)
\(\Rightarrow (k^{\frac{1}{y}})^2 = k^{\frac{1}{x}} \times k^{\frac{1}{z}} \)
\(\Rightarrow k^{\frac{2}{y}} = k^{(\frac{1}{x} + \frac{1}{z})}\)
Therefore, \(\frac{2}{y} = \frac{(x+z)}{xz} \)
\(\Rightarrow \frac{y}{2} = \frac{xz}{(x+z)} \)
\(\Rightarrow y = \frac{2xz}{(x + z)}\)
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If \(4^{4m+2} = 8^{6m-4}\), solve for \(m\)
Answer: D
Exp: \(4^{4m+2} = (2^3)^{6m-4} \)
\(\Rightarrow 4^{4m+2} = 2^{18m+12}\)
Equating powers of 2 both sides,
\(4m + 2 = 18m -12 \)
\(\Rightarrow 14=14m \)
\(\Rightarrow m=1\)
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