If 100!*100!../99!99!... , then what is the value of x ?

Question

If  \frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt {10^x} , then what is the value of x ?

A. 6
B. 7
C. 8
D. 12
E. 14

Bunuel · Accepted Answer

Official Solution:

If  \frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt {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 {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 {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

gmatophobia · Answer

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

Attachments

Screenshot 2022-10-03 165026.png [ 153.57 KiB | Viewed 8030 times ]

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

SaquibHGMATWhiz · Answer

Solution:

We have  \frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt {10^x}

⇒\frac{\sqrt{100	imes 99!\sqrt{100	imes 99!\sqrt{100	imes 99!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt {10^x}

⇒\frac{10\sqrt{99!	imes 10\sqrt{99!	imes 10\sqrt{99!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt {10^x}

⇒\frac{10	imes 10^{\frac{1}{2}}	imes 10^{\frac{1}{4}}\sqrt{99!\sqrt{99!\sqrt{99!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt {10^x}

⇒10^{1+\frac{1}{2}+\frac{1}{4}}=\sqrt {10^x}

⇒10^{\frac{8+4+2}{8}}=10^{\frac{x}{8}}

⇒x=8+4+2=14

Hence the right answer is    Option E

Vetrick · Answer

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.

Regor60 · Answer

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

Fdambro294 · Answer

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

^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:  ^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 · Answer

If  \frac{\sqrt{100!\sqrt{100!\sqrt{100!}}}}{\sqrt{99!\sqrt{99!\sqrt{99!}}}}=\sqrt {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

If 100!*100!../99!99!... , then what is the value of x ?

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If 100!*100!../99!99!... , then what is the value of x ?

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