4. What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10


a(r^n - 1) / (r-1)
5+4*5+4*5^2+4*5^3+4*5^4+4*5^5
1+4+4*5+4*5^2+4*5^3+4*5^4+4*5^5
1 + 4(5^6 -1)/4
5^6

Option

6. If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32


x^5 = -37

-2^5 = -32
-3<x<-2

A. Eliminated
B. Eliminated
C. x^2<4 = -2<x<2 Eliminated
D. x^3<-8 Correct


Option

10. Given that 5x=125-3y+z and \sqrt{5x}-25-\sqrt{z-3y}=0, then what is the value of \sqrt{\frac{45(z-3y)}{x}}?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

5x-125=z-3y

sqrt{5x}-25=\sqrt{z-3y}
(sqrt{5x}-25)^2=z-3y

(sqrt{5x}-25)^2=5x-125
\sqrt{5x}=15
x=5*9=45

so the expression \sqrt{\frac{45(z-3y)}{x} = \sqrt{45(15-25)/45}=\sqrt{-10}

E is the answ
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9. If x is a positive integer then the value of \frac{22^{22x}-22^{2x}}{11^{11x}-11^x} is closest to which of the following?
A. 2^{11x}
B. 11^{11x}
C. 22^{11x}
D. 2^{22x}*11^{11x}
E. 2^{22x}*11^{22x}


let 11=a

so we have

(2a^(2ax)-2a^2x)/(a^(ax)-a^x)=2*((a^(ax)-a^x)*(a^(ax)+a^x))/(a^(ax)-a^x)=2(a^(ax)+a^x)

2(a^(ax)+a^x)=2*(11^(11x)+11^x)
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And it looks as though every other question has been solved except for this one. I won't solve it completely, but I'll help people get started. Here, since we are asked for an approximation only, we don't need to compute an exact value. You might notice that if x is a positive integer, \(22^{22x}\) is vastly bigger than \(22^{2x}\). Since that's true, if we only need to estimate, we can ignore the small term. We can do the same thing in the denominator.
_________________

2a^(2ax)/a^(ax)=2*a^(ax)=2*11^(11x)

ans is ?
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Hello Ian. Glad to be back!

Thanks for your comments. Yes, there were 2 typos:
For Q5: included the exponent 26 in braces {26} to avoid confusion;
For Q3: added the brackets which were lost in copy/past, it should read \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\).
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i believe the answer is : 7^12-3^6 ---> 11-9= 2 ---> B

note: I believe the part I wrote (which i colored red) is wrong. Can someone explain why? when an exponent is raised to another exponent then what operation should we do first? - Okey! Just read the rules about routs that Bunuel posted and found my mistake! (exponentiation by stacked symbols)

The answer should be E. 14

Here is how:

\(5^{10x}=4900\)
so \({(5^{5x})}^2=70^2\)

Take square root of both sides >

so \(5^{5x}=70\)

Also from question stem:

\(2^{\sqrt{y}} = 5^2\)

So (from the question stem)

\(4^{-\sqrt{y}} = \frac{1}{4^{\sqrt{y}}}\)
\(4^{-\sqrt{y}} = (\frac{1}{2^{\sqrt{y}}})^2 = \frac{1}{5^4}\)

Now \(5^{(x-1)^5}\)

Should be \(\frac{5^{5x}}{5^5}\)

We know \(5^{10x}=4900\)

so \(5^{10x}=70^2\)

so \(5^{5x}=70\)

So \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)

is basically \(\frac{5^x}{5^5}*5^4\)

We already know the value of \(5^x\) which is \(70\)

So now it becomes \(\frac{70}{5^5}*5^4\)

Which should resolve to \(14\)

Hence Answer = E = \(14\)

Q2. What is the units digit of \((17^3)^4-1973^{3^2}\)?[/b]

Now we know that:

\((17^3)^4 = (17^6)^2\)

So the expression now becomes

\((17^6)^2 - (1973^3)^2\)

Now we know that \(a^2-b^2\) can be written as \((a+b)*(a-b)\)

Let \(a=17^6\) and \(b=1973^3\)

Now on to \((a+b)*(a-b)\)

\((17^6+1973^3)*(17^6-1973^3)\)

Lets see what the units digit of \(17^6\) is:
We are only concerned with \(7\):
\(7*7*7*7*7*7= ..........9\) unit digit of 9

Lets see what the units digit of \(1973^3\) is:
We are only concerned with \(3\):
\(3*3*3= ..........7\) unit digit of 7

Lets translate this into our expression for \((a+b)*(a-b)\)

\((17^6+1973^3)*(17^6-1973^3)\)

Should in digit terms give:

\((9+7)*(9-7)\)
\((6)*(2)\)

Should Yield a units digit of 2.

Hence Answer = 2