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$\begin{aligned} If \log_{10} a = p, \log_{10} b = q, \\ then what is \log_{10} (a^{p} b^{q}) equal to ? \end{aligned}$ $\eqalign{ & {\text{If}}\,{\log _{10}}a = p,\,\,\,{\log _{10}}b = q, \cr & {\text{then}}\,{\text{what}}\,\,{\text{is}}\,\,{\log _{10}}\left( {{a^p}{b^q}} \right)\,\,{\text{equal}}\,\,{\text{to}}? \cr}$

1p² + q²
2p² - q²
3p²q²
4^p²/_q²

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given, \\ \log_{10} a = p, \log_{10} b = q \\ \log_{10} (a^{p} b^{q}) = \log_{10} a^{p} + \log_{10} b^{q} \\ = p \log_{10} a + q \log_{10} b \\ = p^{2} + q^{2} \end{aligned}

$\eqalign{ & {\text{Given}}, \cr & {\log _{10}}a = p,\,{\log _{10}}b = q \cr & {\log _{10}}\left( {{a^p}{b^q}} \right) = {\log _{10}}{a^p} + {\log _{10}}{b^q} \cr & = p{\log _{10}}a + q{\log _{10}}b \cr & = {p^2} + {q^2} \cr}$

$\begin{aligned} If \log 3 \log (3^{x} - 2) and \log (3^{x} + 4) \\ are in arithmetic progression, \\ then x is equal to - \end{aligned}$ $\eqalign{ & {\text{If}}\,\log 3\log \left( {{3^x} - 2} \right)\,{\text{and}}\,\log \left( {{3^x} + 4} \right)\, \cr & {\text{are}}\,{\text{in}}\,{\text{arithmetic}}\,\,{\text{progression}}\,{\text{,}} \cr & {\text{then}}\,x\,{\text{is}}\,{\text{equal}}\,{\text{to}} - \cr}$

1⁸/₃
2log 3⁸
3log 2³
48

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} In arithmetic progression common ratio are equal to \\ \log (3^{x} - 2) - \log 3 \\ = \log (3^{x} + 4) - \log (3^{x} - 2) \\ = \frac{\log (3^{x} - 2)}{\log 3} \\ = \frac{\log (3^{x} + 4)}{\log (3^{x} - 2)} (∴ \log a - \log b = \log \frac{a}{b}) \\ = \frac{\log 3^{x}}{\log 2 \log 3} \\ = \frac{x \log 3 \log 4 \log 2}{x \log 3} \\ = \frac{x \log 3}{\log 2 \log 3} \\ = \frac{x \log 3 \log 4 \log 2}{x \log 2} \\ = \frac{x}{\log 2} \\ = \log 4 \log 2 \\ \Rightarrow x = \log 4 \log 2 \log 2 \\ \Rightarrow x = \log 8 \\ \Rightarrow x = \log 2^{3} \end{aligned}

$\eqalign{ & {\text{In arithmetic progression common ratio are equal to}} \cr & \log \left( {{3^x} - 2} \right) - \log 3 \cr & = \log \left( {{3^x} + 4} \right) - \log \left( {{3^x} - 2} \right) \cr & = \frac{{\log \left( {{3^x} - 2} \right)}}{{\log 3}} \cr & = \frac{{\log \left( {{3^x} + 4} \right)}}{{\log \left( {{3^x} - 2} \right)}}\left( {\therefore \log a - \log b = \log \frac{a}{b}} \right) \cr & = \frac{{\log {3^x}}}{{\log 2\log 3}} \cr & = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & = \frac{{x\log 3}}{{\log 2\log 3}} \cr & = \frac{{x\log 3\log 4\log 2}}{{x\log 2}} \cr & = \frac{x}{{\log 2}} \cr & = \log 4\log 2 \cr & \Rightarrow x = \log 4\log \,2\log 2 \cr & \Rightarrow x = \log 8 \cr & \Rightarrow x = \log {2^3} \cr}$

$\begin{aligned} The number of digits in 4^{9} \times 5^{17}, \\ when expressed in usual form, is - \end{aligned}$ $\eqalign{ & {\text{The number of digits in }}{{\text{4}}^9} \times {{\text{5}}^{17}}{\text{,}} \cr & {\text{when expressed in usual form, is - }} \cr}$

116
217
318
419

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log (4^{9} \times 5^{17}) \\ = \log (4^{9}) + \log (5^{17}) \\ = \log (2^{2}) 9 + \log (5^{17}) \\ = \log (2^{18}) + \log (5^{17}) \\ = 18 \log 2 + 17 \log 5 \\ = 18 \log 2 + 17 (\log 10 - \log 2) \\ = 18 \log 2 + 17 \log 10 - 17 \log 2 \\ = \log 2 + 17 \log 10 \\ = 0.3010 + 17 \times 1 = 17.3010 \\ ∴ characteristic = 17, \\ Hence, the number of digits in (4^{9} \times 5^{17}) = 18 \end{aligned}

$\eqalign{ & \log \left( {{4^9} \times {5^{17}}} \right) \cr & = \log \left( {{4^9}} \right) + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^2}} \right)9 + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^{18}}} \right) + \log \left( {{5^{17}}} \right) \cr & = 18\log 2 + 17\log 5 \cr & = 18\log 2 + 17\left( {\log 10 - \log 2} \right) \cr & = 18\log 2 + 17\log 10 - 17\log 2 \cr & = \log 2 + 17\log 10 \cr & = 0.3010 + 17 \times 1 = 17.3010 \cr & \therefore {\text{characteristic}} = 17, \cr & {\text{Hence, the number of digits in }}\left( {{4^9} \times {5^{17}}} \right) = 18 \cr}$

$\begin{aligned} If \log 2 = 0.30103, \\ the number of digits in 4^{50} is - \end{aligned}$ $\eqalign{ & {\text{If}}\log 2 = 0.30103, \cr & {\text{the}}\,{\text{number}}\,{\text{of}}\,{\text{digits}}\,{\text{in}}\,{4^{50}}{\text{is}} - \cr}$

130
231
3100
4200

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log 4^{0} \\ = 50 \log 4 \\ = 50 \log 2^{2} \\ = (50 \times 2) \log 2 \\ = 100 \times \log 2 \\ = (100 \times 0.30103) \\ = 30.103 \\ ∴ characteristic = 30, \\ Hence, the number of digits in 4^{50} = 31 \end{aligned}

$\eqalign{ & \log {4^0} \cr & = 50\log 4 \cr & = 50\log {2^2} \cr & = \left( {50 \times 2} \right)\log 2 \cr & = 100 \times \log 2 \cr & = \left( {100 \times 0.30103} \right) \cr & = 30.103 \cr & \therefore {\text{characteristic}} = 30, \cr & {\text{Hence, the number of digits in }}{{\text{4}}^{50}} = 31 \cr}$

$\begin{aligned} If \log 2 = 0.3010 a n d \log 3 = 0.4771, \\ t h e v a l u e o f \log_{5} 512 = ? \end{aligned}$ $\eqalign{ & {\text{If}}\,\log 2 = 0.3010\,and\,\log 3 = 0.4771, \cr & the\,value\,of\,{\log _5}512 = ? \cr}$

12.870
22.967
33.876
43.912

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{5} 512 \\ = \frac{\log 512}{\log 5} \\ = \frac{\log 2^{9}}{\log (\frac{10}{2})} \\ = \frac{9 \log 2}{\log 10 - \log 2} \\ = \frac{(9 \times 0.3010)}{1 - 0.3010} \\ = \frac{2.709}{0.699} \\ = \frac{2709}{699} \\ = 3.876 \end{aligned}

$\eqalign{ & {\log _5}512 \cr & = \frac{{\log 512}}{{\log 5}} \cr & = \frac{{\log {2^9}}}{{\log \left( {\frac{{10}}{2}} \right)}} \cr & = \frac{{9\log 2}}{{\log 10 - \log 2}} \cr & = \frac{{\left( {9 \times 0.3010} \right)}}{{1 - 0.3010}} \cr & = \frac{{2.709}}{{0.699}} \cr & = \frac{{2709}}{{699}} \cr & = 3.876 \cr}$

$\begin{aligned} If a = lo g_{8} 225 and b = lo g_{2} 15, \\ then a in terms of b is - \end{aligned}$ $\eqalign{ & {\text{If}}\,{\text{a = lo}}{{\text{g}}_{\text{8}}}{\text{225}}\,{\text{and}}\,{\text{b = lo}}{{\text{g}}_{\text{2}}}{\text{15}}, \cr & {\text{then}}\,a\,{\text{in}}\,{\text{terms}}\,{\text{of}}\,b\,{\text{is}} - \cr}$

1^b/₂
2^2b/₃
3b
4^3b/₂

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a = \log_{8} 225 \\ = \log_{2^{3}} (15^{2}) \\ = \frac{2}{3} \log_{2} 15 \\ = \frac{2 b}{3} \end{aligned}

$\eqalign{ & a = {\log _8}225 \cr & = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr & = \frac{2}{3}{\log _2}15 \cr & = \frac{{2b}}{3} \cr}$

$\begin{aligned} If \log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0, \\ what is the value of x ? \end{aligned}$ $\eqalign{ & {\text{If}}\,\,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0, \cr & {\text{what}}\,\,{\text{is}}\,\,{\text{the}}\,\,{\text{value}}\,\,{\text{of}}\,\,x? \cr}$

12
23
34
45

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0 \\ \Rightarrow \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 7^{0} = 1 \\ \Rightarrow \sqrt{x + 5} + \sqrt{x} = 5^{1} = 5 \\ \Rightarrow {(\sqrt{x + 5} + \sqrt{x})}^{2} = 25 \\ \Rightarrow (x + 5) + x + 2 \sqrt{x + 5} \sqrt{x} = 25 \\ \Rightarrow 2 x + 2 \sqrt{x} \sqrt{x + 5} = 20 \\ \Rightarrow \sqrt{x} \sqrt{x + 5} = 10 \\ \Rightarrow x (x + 5) = {(10 - x)}^{2} \\ \Rightarrow x^{2} + 5 x = 100 + x^{2} - 20 x \\ \Rightarrow 25 x = 100 \\ \Rightarrow x = 4 \end{aligned}

$\eqalign{ & \,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0 \cr & \Rightarrow {\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = {7^0} = 1 \cr & \Rightarrow \sqrt {x + 5} + \sqrt x = {5^1} = 5 \cr & \Rightarrow {\left( {\sqrt {x + 5} + \sqrt x } \right)^2} = 25 \cr & \Rightarrow \left( {x + 5} \right) + x + 2\sqrt {x + 5} \sqrt x = 25 \cr & \Rightarrow 2x + 2\sqrt x \sqrt {x + 5} = 20 \cr & \Rightarrow \sqrt x \sqrt {x + 5} = 10 \cr & \Rightarrow x\left( {x + 5} \right) = {\left( {10 - x} \right)^2} \cr & \Rightarrow {x^2} + 5x = 100 + {x^2} - 20x \cr & \Rightarrow 25x = 100 \cr & \Rightarrow x = 4 \cr}$

$\begin{aligned} If \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9, \\ then x equals - \end{aligned}$ $\eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr}$

13
29
327
4None of these

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9 \\ \Rightarrow \log_{3} x + \log_{32} x^{2} + \log_{33} x^{3} = 9 \\ \Rightarrow \log_{3} x + \frac{2}{2} \log_{3} x + \frac{3}{3} \log_{3} x = 9 \\ \Rightarrow 3 \log_{3} x = 9 \\ \Rightarrow \log_{3} x = 3 \\ \Rightarrow x = 3^{3} = 27 \end{aligned}

$\eqalign{ & {\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{32}}{x^2} + {\log _{33}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr}$

$\begin{aligned} If \log x - 5 \log 3 = - 2, \\ then x equals - \end{aligned}$ $\eqalign{ & {\text{If}}\,\,\log x - 5\log 3 = - 2, \cr & {\text{then }}x{\text{ equals - }} \cr}$

10.8
20.81
31.25
42.43

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log x - 5 \log 3 = - 2 \\ \Rightarrow \log x - \log 3^{5} = - 2 \\ \Rightarrow \log (\frac{x}{3^{5}}) = - 2 \\ \Rightarrow \frac{x}{243} = 10^{- 2} = \frac{1}{100} \\ \Rightarrow x = \frac{243}{100} = 2.43 \end{aligned}

$\eqalign{ & \log x - 5\log 3 = - 2 \cr & \Rightarrow \log x - \log {3^5} = - 2 \cr & \Rightarrow \log \left( {\frac{x}{{{3^5}}}} \right) = - 2 \cr & \Rightarrow \frac{x}{{243}} = {10^{ - 2}} = \frac{1}{{100}} \cr & \Rightarrow x = \frac{{243}}{{100}} = 2.43 \cr}$

$\begin{aligned} If \log_{10} 7 = a, then \log_{10} (\frac{1}{70}) \\ is equal to - \end{aligned}$ $\eqalign{ & {\text{If}}\,\,{\log _{10}}7 = a,{\text{then}}\,\,{\log _{10}}\left( {\frac{1}{{70}}} \right)\,\, \cr & {\text{is}}\,\,{\text{equal}}\,\,{\text{to}} - \cr}$

1-(1 + a)
2(1 + a)^-1
3^a/₁₀
4¹/_10a

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{10} (\frac{1}{70}) \\ = \log_{10} 1 - \log_{10} 70 \\ = - \log_{10} (7 \times 10) \\ = - (\log_{10} 7 + \log_{10} 10) \\ = - (a + 1) \end{aligned}

$\eqalign{ & {\log _{10}}\left( {\frac{1}{{70}}} \right) \cr & = {\log _{10}}1 - {\log _{10}}70 \cr & = - {\log _{10}}\left( {7 \times 10} \right) \cr & = - \left( {{{\log }_{10}}7 + {{\log }_{10}}10} \right) \cr & = - \left( {a + 1} \right) \cr}$

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Logarithm - 01

$\begin{aligned} The number of digits in 4^{9} \times 5^{17}, \\ when expressed in usual form, is - \end{aligned}$ $\eqalign{ & {\text{The number of digits in }}{{\text{4}}^9} \times {{\text{5}}^{17}}{\text{,}} \cr & {\text{when expressed in usual form, is - }} \cr}$

$\begin{aligned} If \log 2 = 0.30103, \\ the number of digits in 4^{50} is - \end{aligned}$ $\eqalign{ & {\text{If}}\log 2 = 0.30103, \cr & {\text{the}}\,{\text{number}}\,{\text{of}}\,{\text{digits}}\,{\text{in}}\,{4^{50}}{\text{is}} - \cr}$

$\begin{aligned} If \log 2 = 0.3010 a n d \log 3 = 0.4771, \\ t h e v a l u e o f \log_{5} 512 = ? \end{aligned}$ $\eqalign{ & {\text{If}}\,\log 2 = 0.3010\,and\,\log 3 = 0.4771, \cr & the\,value\,of\,{\log _5}512 = ? \cr}$

$\begin{aligned} If \log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0, \\ what is the value of x ? \end{aligned}$ $\eqalign{ & {\text{If}}\,\,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0, \cr & {\text{what}}\,\,{\text{is}}\,\,{\text{the}}\,\,{\text{value}}\,\,{\text{of}}\,\,x? \cr}$

$\begin{aligned} If \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9, \\ then x equals - \end{aligned}$ $\eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr}$

$\begin{aligned} If \log x - 5 \log 3 = - 2, \\ then x equals - \end{aligned}$ $\eqalign{ & {\text{If}}\,\,\log x - 5\log 3 = - 2, \cr & {\text{then }}x{\text{ equals - }} \cr}$

$\begin{aligned} If \log_{10} 7 = a, then \log_{10} (\frac{1}{70}) \\ is equal to - \end{aligned}$ $\eqalign{ & {\text{If}}\,\,{\log _{10}}7 = a,{\text{then}}\,\,{\log _{10}}\left( {\frac{1}{{70}}} \right)\,\, \cr & {\text{is}}\,\,{\text{equal}}\,\,{\text{to}} - \cr}$

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Logarithm - 01

03. The number of digits in 49×517,when expressed in usual form, is - The number of digits in 49×517,when expressed in usual form, is - \eqalign{ & {\text{The number of digits in }}{{\text{4}}^9} \times {{\text{5}}^{17}}{\text{,}} \cr & {\text{when expressed in usual form, is - }} \cr}

04. Iflog2=0.30103,thenumberofdigitsin450is−Iflog⁡2=0.30103,thenumberofdigitsin450is−\eqalign{ & {\text{If}}\log 2 = 0.30103, \cr & {\text{the}}\,{\text{number}}\,{\text{of}}\,{\text{digits}}\,{\text{in}}\,{4^{50}}{\text{is}} - \cr}

05. Iflog2=0.3010andlog3=0.4771,thevalueoflog5512=?Iflog⁡2=0.3010andlog⁡3=0.4771,thevalueoflog5512=?\eqalign{ & {\text{If}}\,\log 2 = 0.3010\,and\,\log 3 = 0.4771, \cr & the\,value\,of\,{\log _5}512 = ? \cr}

07. Iflog7log5(x+5−−−−−√+x−−√)=0,whatisthevalueofx?Iflog7log5(x+5+x)=0,whatisthevalueofx?\eqalign{ & {\text{If}}\,\,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0, \cr & {\text{what}}\,\,{\text{is}}\,\,{\text{the}}\,\,{\text{value}}\,\,{\text{of}}\,\,x? \cr}

08. Iflog3x+log9–x2+log27x3=9,thenxequals - Iflog3x+log9_x2+log27x3=9,thenxequals - \eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr}

09. Iflogx−5log3=−2,then x equals - Iflog⁡x−5log⁡3=−2,then x equals - \eqalign{ & {\text{If}}\,\,\log x - 5\log 3 = - 2, \cr & {\text{then }}x{\text{ equals - }} \cr}

10. Iflog107=a,thenlog10(170)isequalto−Iflog107=a,thenlog10(170)isequalto−\eqalign{ & {\text{If}}\,\,{\log _{10}}7 = a,{\text{then}}\,\,{\log _{10}}\left( {\frac{1}{{70}}} \right)\,\, \cr & {\text{is}}\,\,{\text{equal}}\,\,{\text{to}} - \cr}

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NCERT 6 to 12

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$\begin{aligned} The number of digits in 4^{9} \times 5^{17}, \\ when expressed in usual form, is - \end{aligned}$ $\eqalign{ & {\text{The number of digits in }}{{\text{4}}^9} \times {{\text{5}}^{17}}{\text{,}} \cr & {\text{when expressed in usual form, is - }} \cr}$

$\begin{aligned} If \log 2 = 0.30103, \\ the number of digits in 4^{50} is - \end{aligned}$ $\eqalign{ & {\text{If}}\log 2 = 0.30103, \cr & {\text{the}}\,{\text{number}}\,{\text{of}}\,{\text{digits}}\,{\text{in}}\,{4^{50}}{\text{is}} - \cr}$

$\begin{aligned} If \log 2 = 0.3010 a n d \log 3 = 0.4771, \\ t h e v a l u e o f \log_{5} 512 = ? \end{aligned}$ $\eqalign{ & {\text{If}}\,\log 2 = 0.3010\,and\,\log 3 = 0.4771, \cr & the\,value\,of\,{\log _5}512 = ? \cr}$

$\begin{aligned} If \log_{3} x + \log_{\underline{9}} x^{2} + \log_{27} x^{3} = 9, \\ then x equals - \end{aligned}$ $\eqalign{ & {\text{If}}\,{\log _3}x + {\log _{\underline 9 }}{x^2} + {\log _{27}}{x^3} = 9, \cr & {\text{then}}\,\,x\,\,{\text{equals - }} \cr}$

$\begin{aligned} If \log x - 5 \log 3 = - 2, \\ then x equals - \end{aligned}$ $\eqalign{ & {\text{If}}\,\,\log x - 5\log 3 = - 2, \cr & {\text{then }}x{\text{ equals - }} \cr}$

$\begin{aligned} If \log_{10} 7 = a, then \log_{10} (\frac{1}{70}) \\ is equal to - \end{aligned}$ $\eqalign{ & {\text{If}}\,\,{\log _{10}}7 = a,{\text{then}}\,\,{\log _{10}}\left( {\frac{1}{{70}}} \right)\,\, \cr & {\text{is}}\,\,{\text{equal}}\,\,{\text{to}} - \cr}$

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