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Difficulty:
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Question Stats:
60% (02:24) correct
40%
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wrong
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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
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A. 6
B. 7
C. 8
D. 12
E. 14
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Bunuel
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
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General Discussion
gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
We can simplify the denominator first and then use that information to work the numerator.
Denominator = \(\sqrt{99!\sqrt{99!\sqrt{99!}}}\)
Assume
\(\sqrt{99!}\) = x
\(\sqrt{99! * x}\) = y
\(\sqrt{99! * y}\) = z
\(\sqrt{99!\sqrt{99!\sqrt{99!}}}\) = z
Thus the denominator can be denoted by z
Numerator= \(\sqrt{100!\sqrt{100!\sqrt{100!}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99! * 100}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99!}* \sqrt{100}}}\)
\(\sqrt{100!\sqrt{100! * x * 10}}\)
\(\sqrt{100!\sqrt{100 * 99! * x * 10}}\)
\(\sqrt{100!\sqrt{1000 * 99! * x}}\)
\(\sqrt{100! * 10^{\frac{3}{2}} * Y }\)
\(\sqrt{10^{7/2} * 99! * y }\)
\(10^{7/4} * z\)
LHS = \(10^{7/4} * \frac{z }{ z}\) = \(10^{7/4}\)
RHS = \(10^{x/8}\)
Comparing
\(\frac{x}{8}\) = \(\frac{7}{4}\)
x = 14
Method 2 - In fact an easier one
Raise both sides to the power of 8
\(\frac{100!^{4} * 100!^{2} * 100! }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(\frac{100^{4} * 100^{2} * 100 * (99!^{4} * 99!^{2} * 99!) }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(10^{14}\) = \(10^x\)
x = 14
IMO - E
_________________
Denominator = \(\sqrt{99!\sqrt{99!\sqrt{99!}}}\)
Assume
\(\sqrt{99!}\) = x
\(\sqrt{99! * x}\) = y
\(\sqrt{99! * y}\) = z
\(\sqrt{99!\sqrt{99!\sqrt{99!}}}\) = z
Thus the denominator can be denoted by z
Numerator= \(\sqrt{100!\sqrt{100!\sqrt{100!}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99! * 100}}}\)
\(\sqrt{100!\sqrt{100! * \sqrt{99!}* \sqrt{100}}}\)
\(\sqrt{100!\sqrt{100! * x * 10}}\)
\(\sqrt{100!\sqrt{100 * 99! * x * 10}}\)
\(\sqrt{100!\sqrt{1000 * 99! * x}}\)
\(\sqrt{100! * 10^{\frac{3}{2}} * Y }\)
\(\sqrt{10^{7/2} * 99! * y }\)
\(10^{7/4} * z\)
LHS = \(10^{7/4} * \frac{z }{ z}\) = \(10^{7/4}\)
RHS = \(10^{x/8}\)
Comparing
\(\frac{x}{8}\) = \(\frac{7}{4}\)
x = 14
Method 2 - In fact an easier one
Raise both sides to the power of 8
\(\frac{100!^{4} * 100!^{2} * 100! }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(\frac{100^{4} * 100^{2} * 100 * (99!^{4} * 99!^{2} * 99!) }{ 99!^{4} * 99!^{2} * 99!}\) = \(10^x\)
\(10^{14}\) = \(10^x\)
x = 14
IMO - E
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
GMAT 1: 760 Q51 V40
Posts: 623
Bunuel
Solution:
We have \(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\)
\(⇒\frac{\sqrt{100\times 99!\sqrt{100\times 99!\sqrt{100\times 99!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\)
\(⇒\frac{10\sqrt{99!\times 10\sqrt{99!\times 10\sqrt{99!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\)
\(⇒\frac{10\times 10^{\frac{1}{2}}\times 10^{\frac{1}{4}}\sqrt{99!\sqrt{99!\sqrt{99!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt[8]{10^x}\)
\(⇒10^{1+\frac{1}{2}+\frac{1}{4}}=\sqrt[8]{10^x}\)
\(⇒10^{\frac{8+4+2}{8}}=10^{\frac{x}{8}}\)
\(⇒x=8+4+2=14\)
Hence the right answer is Option E
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The right most 100! & the right most 99! have been square-rooted thrice. So, they are (100!)^1/8 & (99!)^1/8
(1/2 * 1/2 * 1/2 = 1/8).
The 100! and the 99! in the center are (100!)^1/4 & (99!)^1/4 -> for they have been square-rooted only twice.
The left most 100! and the 99! are (100!)^1/2 & (99!)^1/2 -> for they have been square-rooted only once.
so the numerator is (100!)^1/2 * (100!)^1/4 * (100!)^1/8 which is (100!)^7/8 for ((1/2)+(1/4)+(1/8))=7/8
Similarly the denominator is (99!)^7/8.
Now LHS of the eqn is (100!)^7/8 / (99!)^7/8
If each number in the factorials carries the power 7/8, all the numbers in the numerator gets cancelled out except 100^(7/8). Therefore LHS is reduced to 100^(7/8).
Now, RHS is 10^(x/8). To make LHS also 10 to the power of, rewrite LHS as (10^2)^7/8 which can simplified to 10^(14/8)
Comparing LHS to RHS, X = 14.
So, answer is E.
(1/2 * 1/2 * 1/2 = 1/8).
The 100! and the 99! in the center are (100!)^1/4 & (99!)^1/4 -> for they have been square-rooted only twice.
The left most 100! and the 99! are (100!)^1/2 & (99!)^1/2 -> for they have been square-rooted only once.
so the numerator is (100!)^1/2 * (100!)^1/4 * (100!)^1/8 which is (100!)^7/8 for ((1/2)+(1/4)+(1/8))=7/8
Similarly the denominator is (99!)^7/8.
Now LHS of the eqn is (100!)^7/8 / (99!)^7/8
If each number in the factorials carries the power 7/8, all the numbers in the numerator gets cancelled out except 100^(7/8). Therefore LHS is reduced to 100^(7/8).
Now, RHS is 10^(x/8). To make LHS also 10 to the power of, rewrite LHS as (10^2)^7/8 which can simplified to 10^(14/8)
Comparing LHS to RHS, X = 14.
So, answer is E.
Bunuel
100!/99! is simply 100=100*99*.../99*...
So the innermost root is 10, but then it's raised to the 1/4 power,so
10^1/4
The next root also 10 is raised to the 1/2 power,so
10^1/2
The final 10 is already out of the root, so
10. Therefore
10^.25 * 10^.5 * 10 =
10^(7/4) = 10^(X/8), so
7/4 = X/8 and X=14
Posted from my mobile device
Posts: 1,324
First, set up the root on the right hand side of the equation as a Fractional Exponent:
(10)^ (x/8)
Step 1: square both sides —- has the effect of removing the outer Root on the left
[ (10)^(x/8) ]^2 = (10)^(2x/8)
And
100! / 99! = 100
Step 2: square both sides again —- now, we must remember to remove the 2nd root AND square the (100) factor that we got after canceling the Factorials
Right side becomes: [ (10)^(2x/8) ]^2 =
(10)^(4x/8)
And left side we have:
(100)^2 ——- and again, the factorials that just opened — (100! / 99!) — will cancel and give us p 100
(10)^(4x/8) = (100)^2 * 100 * sqrt( 100! / 99! )
At this point we can just cancel the factorials within the square root and take the square root of (100)
(10)^(4x/8) = (100)^2 * 100 * sqrt(100)
(10)^(x / 2) = (100)^2 * 100 * 10
Set up all the terms as a Base of (10)
(10)^(x/2) = (10)^4 * (10)^2 * (10)^1
(10)^(x/2) = (10)^7
Drop the Base and set the Exponents equal:
(x/2) = 7
X = 14
*E*
Posted from my mobile device
(10)^ (x/8)
Step 1: square both sides —- has the effect of removing the outer Root on the left
[ (10)^(x/8) ]^2 = (10)^(2x/8)
And
100! / 99! = 100
Step 2: square both sides again —- now, we must remember to remove the 2nd root AND square the (100) factor that we got after canceling the Factorials
Right side becomes: [ (10)^(2x/8) ]^2 =
(10)^(4x/8)
And left side we have:
(100)^2 ——- and again, the factorials that just opened — (100! / 99!) — will cancel and give us p 100
(10)^(4x/8) = (100)^2 * 100 * sqrt( 100! / 99! )
At this point we can just cancel the factorials within the square root and take the square root of (100)
(10)^(4x/8) = (100)^2 * 100 * sqrt(100)
(10)^(x / 2) = (100)^2 * 100 * 10
Set up all the terms as a Base of (10)
(10)^(x/2) = (10)^4 * (10)^2 * (10)^1
(10)^(x/2) = (10)^7
Drop the Base and set the Exponents equal:
(x/2) = 7
X = 14
*E*
Posted from my mobile device
Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 15 Sep 2026
Posts: 6,129
Location: India
Schools: HBS '28 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
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 ?
\(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=10^{\frac{x}{8}}\)
Squaring both sides, we get
\(\frac{100!\sqrt{100!\sqrt{100!}}}{99!\sqrt{99!\sqrt{99!}}}=10^{\frac{x}{4}}\)
\(\frac{100\sqrt{100!\sqrt{100!}}}{\sqrt{99!\sqrt{99!}}}=10^{\frac{x}{4}}\)
\(\frac{\sqrt{100!\sqrt{100!}}}{\sqrt{99!\sqrt{99!}}}=10^{\frac{x}{4}-2}\)
Squaring both sides, we get
\(\frac{100!\sqrt{100!}}{99!\sqrt{99!}}=10^{\frac{x}{2}-4}\)
\(\frac{100\sqrt{100!}}{\sqrt{99!}}=10^{\frac{x}{2}-4}\)
\(\frac{\sqrt{100!}}{\sqrt{99!}}=10^{\frac{x}{2}-6}\)
Squaring both sides, we get
\(\frac{100!}{99!}=10^{x-12}\)
\(100=10^{x-12}\)
x - 12 = 2
x = 14
IMO E
\(\frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=10^{\frac{x}{8}}\)
Squaring both sides, we get
\(\frac{100!\sqrt{100!\sqrt{100!}}}{99!\sqrt{99!\sqrt{99!}}}=10^{\frac{x}{4}}\)
\(\frac{100\sqrt{100!\sqrt{100!}}}{\sqrt{99!\sqrt{99!}}}=10^{\frac{x}{4}}\)
\(\frac{\sqrt{100!\sqrt{100!}}}{\sqrt{99!\sqrt{99!}}}=10^{\frac{x}{4}-2}\)
Squaring both sides, we get
\(\frac{100!\sqrt{100!}}{99!\sqrt{99!}}=10^{\frac{x}{2}-4}\)
\(\frac{100\sqrt{100!}}{\sqrt{99!}}=10^{\frac{x}{2}-4}\)
\(\frac{\sqrt{100!}}{\sqrt{99!}}=10^{\frac{x}{2}-6}\)
Squaring both sides, we get
\(\frac{100!}{99!}=10^{x-12}\)
\(100=10^{x-12}\)
x - 12 = 2
x = 14
IMO E
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