\(\sqrt{[200\sqrt{[200\sqrt{[200....... \infty ]}]}]} = ?\)
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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If \(5^{a}=3125\), then the value of \(5^{(a-3)}\) is:
Answer: A
\(5^{a}=3125 \Leftrightarrow 5^{a}=5^{5} \)
\(\Rightarrow a = 5 \)
\(\therefore 5^{(a-3)}=5^{(5-3)}=5^{2}=25\)
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\((0.04)^{-1.5}\) = ?
Answer: B
\((0.04)^{-1.5} = (\frac{4}{100})^{-1.5} \)
\(= (\frac{1}{25})^{-\frac{3}{2}} \)
\(= (25)^{\frac{3}{2}} \)
\(= (5^{2})^{\frac{3}{2}} \)
\(= (5)^{2\times \frac{3}{2}} \)
\(= 5^{3} = 125\)
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\(125 \times 125 \times 125 \times 125 \times 125 = 5^?\)
Answer: C
\[ 125 = 5^{3} \quad\Longrightarrow\quad 125^{5} = (5^{3})^{5} = 5^{15} \]
So the exponent is 15. Multiplying five copies of \( 5^{3} \) adds their exponents.
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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 \(x=3+2\sqrt{2}\), then the value of \((\sqrt{x}-\frac{1}{\sqrt{x}})\) is:
Answer: B
\((\sqrt{x}-\frac{1}{\sqrt{x}})^{2}=x+\frac{1}{x}-2\)
\(= (3+2\sqrt{2})+\frac{1}{(3+2\sqrt{2})}-2\)
\(= (3+2\sqrt{2})+\frac{1}{(3+2\sqrt{2})}\times\frac{(3-2\sqrt{2})}{(3-2\sqrt{2})}-2\)
\(= (3+2\sqrt{2})+(3+2\sqrt{2})-2\)
= 4
\(\therefore (\sqrt{x}-\frac{1}{\sqrt{x}})=2\)
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