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21 Feb 2023, 13:36
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If \(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\), then what is the value of x ?
A. \(6\)
B. \(7\)
C. \(8\)
D. \(12\)
E. \(14\)
A. \(6\)
B. \(7\)
C. \(8\)
D. \(12\)
E. \(14\)
21 Feb 2023, 13:36
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Official Solution:
If \(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\), then what is the value of x ?
A. \(6\)
B. \(7\)
C. \(8\)
D. \(12\)
E. \(14\)
Get rid of the roots:
\(\frac { (100!)^{\frac{1}{2}}*(100!)^{\frac{1}{4}}*(100!)^{\frac{1}{8}} } {(99!)^{\frac{1}{2}}*(99!)^{\frac{1}{4}}*(99!)^{\frac{1}{8}} } =\sqrt[8]{10^x}\);
Take the whole expression to the eights power:
\((\frac { (100!)^{\frac{1}{2}}*(100!)^{\frac{1}{4}}*(100!)^{\frac{1}{8}} } {(99!)^{\frac{1}{2}}*(99!)^{\frac{1}{4}}*(99!)^{\frac{1}{8}} } )^8=(\sqrt[8]{10^x})^8\);
\(\frac{(100!)^4*(100!)^2*100!}{(99!)^4*(99!)^2*99!}=10^x\);
\(\frac{(100!)^7}{(99!)^7}=10^x\);
\((\frac{100!}{99!})^7=10^x\);
\(100^7=10^x\);
\(10^{14}=10^x\);
\(x=14\).
Answer: E
If \(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\), then what is the value of x ?
A. \(6\)
B. \(7\)
C. \(8\)
D. \(12\)
E. \(14\)
Get rid of the roots:
\(\frac { (100!)^{\frac{1}{2}}*(100!)^{\frac{1}{4}}*(100!)^{\frac{1}{8}} } {(99!)^{\frac{1}{2}}*(99!)^{\frac{1}{4}}*(99!)^{\frac{1}{8}} } =\sqrt[8]{10^x}\);
Take the whole expression to the eights power:
\((\frac { (100!)^{\frac{1}{2}}*(100!)^{\frac{1}{4}}*(100!)^{\frac{1}{8}} } {(99!)^{\frac{1}{2}}*(99!)^{\frac{1}{4}}*(99!)^{\frac{1}{8}} } )^8=(\sqrt[8]{10^x})^8\);
\(\frac{(100!)^4*(100!)^2*100!}{(99!)^4*(99!)^2*99!}=10^x\);
\(\frac{(100!)^7}{(99!)^7}=10^x\);
\((\frac{100!}{99!})^7=10^x\);
\(100^7=10^x\);
\(10^{14}=10^x\);
\(x=14\).
Answer: E
17 Feb 2025, 22:08
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