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

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

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.

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

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

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