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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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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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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\((256)^{0.16}\times (256)^{0.09}=?\)
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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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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\(\frac{1}{1+x^{(b-a)}+x^{(c-a)}}+\frac{1}{1+x^{(a-b)}+x^{(c-b)}}+\frac{1}{1+x^{(b-c)}+x^{(a-c)}}=?\)
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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\((25)^{7.5}\times(5)^{2.5}\div(125)^{1.5}=5^?\)
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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\((0.04)^{-1.5}=?\)
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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\(\frac{(243)^{n/5}\times 3^{2n+1}}{9^{n}\times 3^{n-1}}=?\)
Answer: C
Given Expression = \(\frac{(243)^{(n/5)}\times 3^{2n+1}}{9^{n}\times 3^{n-1}}\)
\(=\frac{(3^{5})^{(n/5)}\times 3^{2n+1}}{(3^{2})^{n}\times 3^{n-1}}\)
\(=\frac{(3^{5\times(n/5))}\times 3^{2n+1}}{(3^{2n}\times 3^{n-1})}\)
\(=\frac{3^{n\times3^{2n+1}}}{3^{2n}\times3^{n-1}}\)
\(=\frac{3^{(n+2n+1)}}{3^{(2n+n-1)}}\)
\(=\frac{3^{3n+1}}{3^{3n-1}}\)
\(=3^{(3n+1-3n+1)}=3^{2}=9\)
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\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}}=?\)
Answer: B
\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}}=\frac{1}{(1+\frac{a^{n}}{a^{m}})}+\frac{1}{(1+\frac{a^{m}}{a^{n}})}\)
\(=\frac{a^{m}}{(a^{m}+a^{m})}+\frac{a^{m}}{(a^{m}+a^{n})}\)
\(=\frac{(a^{m}+a^{n})}{(a^{m}+a^{n})}\)
=1
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