Solve for \(m\) if \(49(7^m) = 343^{3m+6}\)
Answer: B
Exp: \(49(7^m) = 343^{3m+6} \)
\(\Rightarrow 7^2 7^m = (7^3)^{3m+6} \)
\(\Rightarrow 7^{2+m} = 7^{9m+18}\)
Equating powers of 7 on both sides,
\(\Rightarrow m + 2 = 9m + 18\)
\(\Rightarrow - 16 = 8m\)
\(\Rightarrow m = -2\)
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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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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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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 \(2^x \times 16^{\frac{2}{5}} = 2^{\frac{1}{5}}\), 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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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 \(7^a = 16807\), then the value of \(7^{(a-3)}\) is:
Answer: A
\[ 16807 = 7^{5} \quad\Longrightarrow\quad a = 5 \]
\[ 7^{a-3} = 7^{2} = 49 \]
Equivalently \( \tfrac{16807}{343} = 49 \).
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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 \(2^{(x-y)} = 8\) 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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