\[ \frac{1}{1+a^{(n-m)}} + \frac{1}{1+a^{(m-n)}} = \ ? \]
Answer: C
\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}} \)
\(= \frac{1}{(1+\frac{an}{am})}+\frac{1}{(1+\frac{am}{an})} \)
\(= \frac{am}{(am+an)}+\frac{an}{(am+an)} \)
\(= \frac{(am+an)}{(am+an)} \)
\(= 1\)
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If \(2^x \times 16^{\frac{2}{5}} = 2^{\frac{1}{5}}\), then \(x\) is equal to:
Answer: D
\(2^x \times 16^{\frac{2}{5}} = 2^{\frac{1}{5}}\)
\(\Rightarrow 2^x \times (2^4)^{\frac{2}{5}} = 2^{\frac{1}{5}} \)
\(\Rightarrow 2^x \times 2^{\frac{8}{5}}= 2^{\frac{1}{5}} \)
\(\Rightarrow 2^{(x+ \frac{8}{5})} = 2^{\frac{1}{5}} \)
\(\Rightarrow x + \frac{8}{5} = \frac{1}{5} \)
\(\Rightarrow x = (\frac{1}{5} - \frac{8}{5}) = -\frac{7}{5}\)
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\((25)^{7.5} \times (5)^{2.5} \div (125)^{1.5} = 5?\)
Answer: B
Let \((25)^{7.5} \times (5)^{2.5} \div (125)^{1.5} = 5x.\)
Then, \(\frac{(5)^{7.5} \times (5)^{2.5}}{(5^{3})^{1.5}} = 5x \)
\(\Rightarrow \frac{5^{(2 \times 7.5)} \times (5)^{2.5}}{5^{(3 \times 1.5)}} = 5x\)
\(\Rightarrow \frac{5^{15} \times 5^{2.5}}{5^{4.5}} = 5x\)
\(\Rightarrow 5x = 5^{(15+2.5-4.5)} \)
\(\Rightarrow 5x = 5^{13}, \therefore x = 13\)
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Find the value of \(\left(\dfrac{81}{16}\right)^{-3/4}\).
Answer: B
Correct answer: 8/27
Recognise the base as a fourth power.
\[\frac{81}{16} = \left(\frac{3}{2}\right)^4\]
\[\left(\left(\frac{3}{2}\right)^4\right)^{-3/4} = \left(\frac{3}{2}\right)^{-3} = \left(\frac{2}{3}\right)^3 = \frac{8}{27}\]
The negative sign in the index inverts the fraction; the denominator 4 takes the fourth root and the numerator 3 cubes the result.
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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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\(\frac{1}{1+x^{(b-a)}+x^{(c-a)}}+\frac{1}{1+x^{(a-b)}+x^{(c-b)}}+\frac{1}{1+x^{(b-c)}+x^{(a-c)}}=?\)
Answer: A
Given Exp.= \(\frac{1}{(1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}})}+\frac{1}{(1+\frac{x^{a}}{x^{b}}+\frac{x^{c}}{x^{b}})}+\frac{1}{(1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}})}\)
\(=\frac{x^{a}}{(x^{a}+x^{b}+x^{c})}+\frac{x^{b}}{(x^{a}+x^{b}+x^{c})}+\frac{x^{c}}{(x^{a}+x^{b}+x^{c})}\)
\(=\frac{(x^{a}+x^{b}+x^{c})}{(x^{a}+x^{b}+x^{c})} \)
\(=1\)
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Find the value of \((0.008)^{1/3}\).
Answer: B
Correct answer: 0.2
Write the decimal as a fraction to make the cube root obvious.
\[0.008 = \frac{8}{1000} = \left(\frac{2}{10}\right)^3\]
\[(0.008)^{1/3} = \frac{2}{10} = 0.2\]
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If \(\sqrt{(3+3\sqrt{x})} = 2\), then \(x\) is equal to:
Answer: A
Exp: On squaring both sides, we get:
\(3+3\sqrt{x} = 4\) or \(3\sqrt{x} = 1\)
Cubing both sides, we get \(x = (1\times 1\times 1) = 1\)
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\((25)^{7.5}\times(5)^{2.5}\div(125)^{1.5}=5^?\)
Answer: B
Let \((25)^{7.5}\times(5)^{2.5}\div(125)^{1.5}=5^x\)
Then, \(\frac{(5^{2})^{7.5}\times (5)^{2.5}}{(5^{3})^{1.5}}=5^x\)
\(\Rightarrow \frac{5^{(2\times7.5)}\times5^{2.5}}{(5^{3\times1.5})}=5^x\)
\(\Rightarrow \frac{5^{15}\times5^{2.5}}{5^{4.5}}=5^x\)
\(\Rightarrow 5^{x}=5^{(15+2.5-4.5)}\)
\(\Rightarrow 5^{x}=5^{13}\)
\(\therefore x=13\)
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\((256)^{0.16} \times (256)^{0.09} = ?\)
Answer: A
\((256)^{0.16} \times (256)^{0.09} = (256)^{(0.16+0.09)} \)
\(= (256)^{0.25} \)
\(= (256)^{(25/100)} \)
\(= (256)^{(1/4)} \)
\(= (4^{4})^{(1/4)} \)
\(= 4^{4(1/4)} \)
\(= 4^{1} \)
\(= 4\)
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