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Re: NEW!!! Tough and tricky exponents and roots questions
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14 Jan 2013, 00:38
Thanks Bunuel for these interesting problems.
With respect to problem number 2
"So, we have that the units digit of is 1 and the units digit of is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 1113=2, which gives the final answer that the units digit of is 2."
I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number. And we are trying to find different between smaller number  larger number
eg 757571  58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.
I am unable to think any other way. Please let me know where I am going wrong.
Thanks gmatrant



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Re: NEW!!! Tough and tricky exponents and roots questions
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30 Mar 2013, 14:08
You talk about even and odd in this problem. Wouldn't this not be enough? Since 26 has factors of 2 and 13, even if you had added some even numbers in the product it would still not have sufficed? You would need a 13, 39 etc ? Is my reasoning correct? Bunuel wrote: 5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a nonnegative integer, then what is the value of \(n^{26}26^n\)? A. 26 B. 25 C. 1 D. 0 E. 1
\(23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd\) so \(x\) is an odd number. The only way it to be a multiple of \(26^n\) (even number in integer power) is when \(n=0\), in this case \(26^n=26^0=1\) and 1 is a factor of every integer. Thus \(n=0\) > \(n^{26}26^n=0^{26}26^0=01=1\). Must know for the GMAT: \(a^0=1\), for \(a\neq{0}\)  any nonzero number to the power of 0 is 1. Important note: the case of 0^0 is not tested on the GMAT.
Answer: C.



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Re: NEW!!! Tough and tricky exponents and roots questions
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31 Mar 2013, 08:33
jeremystaub28 wrote: You talk about even and odd in this problem. Wouldn't this not be enough? Since 26 has factors of 2 and 13, even if you had added some even numbers in the product it would still not have sufficed? You would need a 13, 39 etc ? Is my reasoning correct? Bunuel wrote: 5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a nonnegative integer, then what is the value of \(n^{26}26^n\)? A. 26 B. 25 C. 1 D. 0 E. 1
\(23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd\) so \(x\) is an odd number. The only way it to be a multiple of \(26^n\) (even number in integer power) is when \(n=0\), in this case \(26^n=26^0=1\) and 1 is a factor of every integer. Thus \(n=0\) > \(n^{26}26^n=0^{26}26^0=01=1\). Must know for the GMAT: \(a^0=1\), for \(a\neq{0}\)  any nonzero number to the power of 0 is 1. Important note: the case of 0^0 is not tested on the GMAT.
Answer: C. The fact that the product is odd, is already enough to say that it cannot be a multiple of 26^n unless n=0.
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Re: NEW!!! Tough and tricky exponents and roots questions
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16 Apr 2013, 22:53
Bunuel wrote: • \(\sqrt{x^2}=x\), when \(x\leq{0}\), then \(\sqrt{x^2}=x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\)
• When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.
That is, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Even roots have only a positive value on the GMAT.
Hi Bunuel, I am well versed with these exponent rules, however I am becoming confused with the above two rules stated If \(\sqrt{x^2}\) =\(x\) then > \(\sqrt{5^2}\) = \(5\), which means that value can be either 5 or 5. However, as per your example, value should be 5 only. How come this is true? Please clarify. Thanks H



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Re: NEW!!! Tough and tricky exponents and roots questions
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17 Apr 2013, 00:09
imhimanshu wrote: Bunuel wrote: 10. Given that \(5x=1253y+z\) and \(\sqrt{5x}5\sqrt{z3y}=0\), then what is the value of \(\sqrt{\frac{45(z3y)}{x}}\)? A. 5 B. 10 C. 15 D. 20 E. Can not be determined
Hi Bunuel, Request you to please look at my below solution and kindly guide where I am going wrong. Given: 5x = 1253y +z (1) \(\sqrt{5x} = 5+\sqrt{z3y}\) Squaring both sides \(5x= (5+\sqrt{z3y})^2\) \(5x= 125+(z3y) +10\sqrt{(z3y)}\) (2) If you compare equation 1 and 2; then the term \(\sqrt{(z3y)}\) has to be 0 If I put this value in the \(\sqrt{\frac{45(z3y)}{x}}\) Then, the value comes out to be \(\sqrt{0}.\). This is Absurd. I know this is incorrect, but would like to know where I am going wrong. Your help will be appreciated. Thanks H When you square \(5x= (5+\sqrt{z3y})^2\) you should get 5x= 25+10(z3y)^1/2+(z3y) (5^2=25 not 125).
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Re: NEW!!! Tough and tricky exponents and roots questions
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17 Apr 2013, 16:17
Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. Hi Bunuel Can u pls explain the step.....(5^(5x))^2 = 70^2 implies 5^5x = 70 does that mean if a^n = b^n, a=b... Archit



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Re: NEW!!! Tough and tricky exponents and roots questions
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17 Apr 2013, 16:20
Archit143 wrote: Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. Hi Bunuel Can u pls explain the step.....(5^(5x))^2 = 70^2 implies 5^5x = 70 does that mean if a^n = b^n, a=b... Archit It's very simple, take the square root from \((5^{5x})^2=70^2\) to get \(5^{5x}=70\).
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Re: NEW!!! Tough and tricky exponents and roots questions
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20 Jul 2013, 07:44
Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks



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Re: NEW!!! Tough and tricky exponents and roots questions
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20 Jul 2013, 07:50
BankerRUS wrote: Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks Can you please tell me which step is unclear: \(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\).
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Re: NEW!!! Tough and tricky exponents and roots questions
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20 Jul 2013, 08:15
Bunuel wrote: BankerRUS wrote: Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks Can you please tell me which step is unclear: \(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\). The first step with the 4,900 is clear. I do not understand the following: 5^{( 5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14[/m]and how you get to the 70 there.



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Re: NEW!!! Tough and tricky exponents and roots questions
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20 Jul 2013, 09:14
BankerRUS wrote: Bunuel wrote: BankerRUS wrote: Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks
Can you please tell me which step is unclear: \(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\). The first step with the 4,900 is clear. I do not understand the following: 5^{( 5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14[/m]and how you get to the 70 there. \(5^{5x}=70\). So, this step is clear. \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2\). Is this step clear? Next, replace \(5^{5x}\) with 70 and \(2^{\sqrt{y}}\) with 25: \(70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\). Does this make sense?
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Re: NEW!!! Tough and tricky exponents and roots questions
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22 Jul 2013, 09:28
Edit: [edit] [/edit] Bunuel wrote: 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
This question can be solved in several ways:
Traditional approach: \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5)\) Note that we have the sum of geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}1)}{r1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\).
So in our case: \(5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^51)}{51})=5^6\).
30 sec approach based on answer choices: We have the sum of 6 terms. Now, if all terms were equal to the largest term 4*5^5 we would have: \(sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}\), so the actual sum must be less than 5^7, thus the answer must be A: 5^6.
Answer: A. Thanks for the explanation. The part where you take 5+4 out of the full equation is still unclear to me  5+4(5+5^2+5^3+5^4+5^5). Could you please break it down into multiple steps?
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Re: NEW!!! Tough and tricky exponents and roots questions
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22 Jul 2013, 09:39
vaishnogmat wrote: Edit: [edit] [/edit] Bunuel wrote: 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
This question can be solved in several ways:
Traditional approach: \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5)\) Note that we have the sum of geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}1)}{r1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\).
So in our case: \(5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^51)}{51})=5^6\).
30 sec approach based on answer choices: We have the sum of 6 terms. Now, if all terms were equal to the largest term 4*5^5 we would have: \(sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}\), so the actual sum must be less than 5^7, thus the answer must be A: 5^6.
Answer: A. Thanks for the explanation. The part where you take 5+4 out of the full equation is still unclear to me  5+4(5+5^2+5^3+5^4+5^5). Could you please break it down into multiple steps? 4 is factored out from 4*5+4*5^2+4*5^3+4*5^4+4*5^5, not 5+4 from the whole expression: \(4*5+4*5^2+4*5^3+4*5^4+4*5^5=4(5+5^2+5^3+5^4+5^5)\). So, \(5+(4*5+4*5^2+4*5^3+4*5^4+4*5^5)=5+4(5+5^2+5^3+5^4+5^5)\). Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions
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17 Dec 2013, 19:51
Bunuel wrote: 10. Given that \(5x=1253y+z\) and \(\sqrt{5x}5\sqrt{z3y}=0\), then what is the value of \(\sqrt{\frac{45(z3y)}{x}}\)? A. 5 B. 10 C. 15 D. 20 E. Can not be determined
Rearranging both expressions we'll get: \(5x(z3y)=125\) and \(\sqrt{5x}\sqrt{z3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z3y}\) as \(b\).
So we have that \(a^2b^2=125\) and \(ab=5\). Now, \(a^2b^2=(ab)(a+b)=125\) and as \(ab=5\) then \((ab)(a+b)=5*(a+b)=125\) > \(a+b=25\). Thus we get two equations with two unknowns: \(a+b=25\) and \(ab=5\) > solving for \(a\) > \(a=15=\sqrt{5x}\) > \(x=45\). Solving for \(b\) >\(b=10=\sqrt{z3y}\)
Finally, \(\sqrt{\frac{45(z3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).
Answer: B. This is how I went about solving this problem: Given \(\sqrt{5x}  5  \sqrt{z3y}= 0\): Rearrange and square \((\sqrt{5x})^2 = (\sqrt{z3y}+5)^2\) \(5x = z3y + 25 + 10\sqrt{z3y}\) Assuming that x is positive and given \(5x=1253y+z\), set the two equations equal to each other \(125  3y + z = z3y + 25 + 10\sqrt{z3y}\) \(100 = 10\sqrt{z3y}\) \(100 = z3y\)\(5x = 1253y+z\) \(5x = 125 + 100\) \(x = 45\)Thus by substitution: \(= \sqrt{\frac{45(z3y)}{x}}\) \(= \sqrt{\frac{45(100)}{45}}\) \(= \sqrt{100}\) \(= 10\) However, notice that I had to assume that 5x = 5x Why is this permissible in this problem?



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18 Dec 2013, 02:04
NateTheGreat11 wrote: Bunuel wrote: 10. Given that \(5x=1253y+z\) and \(\sqrt{5x}5\sqrt{z3y}=0\), then what is the value of \(\sqrt{\frac{45(z3y)}{x}}\)? A. 5 B. 10 C. 15 D. 20 E. Can not be determined
Rearranging both expressions we'll get: \(5x(z3y)=125\) and \(\sqrt{5x}\sqrt{z3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z3y}\) as \(b\).
So we have that \(a^2b^2=125\) and \(ab=5\). Now, \(a^2b^2=(ab)(a+b)=125\) and as \(ab=5\) then \((ab)(a+b)=5*(a+b)=125\) > \(a+b=25\). Thus we get two equations with two unknowns: \(a+b=25\) and \(ab=5\) > solving for \(a\) > \(a=15=\sqrt{5x}\) > \(x=45\). Solving for \(b\) >\(b=10=\sqrt{z3y}\)
Finally, \(\sqrt{\frac{45(z3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).
Answer: B. This is how I went about solving this problem: Given \(\sqrt{5x}  5  \sqrt{z3y}= 0\): Rearrange and square \((\sqrt{5x})^2 = (\sqrt{z3y}+5)^2\) \(5x = z3y + 25 + 10\sqrt{z3y}\) Assuming that x is positive and given \(5x=1253y+z\), set the two equations equal to each other \(125  3y + z = z3y + 25 + 10\sqrt{z3y}\) \(100 = 10\sqrt{z3y}\) \(100 = z3y\)\(5x = 1253y+z\) \(5x = 125 + 100\) \(x = 45\)Thus by substitution: \(= \sqrt{\frac{45(z3y)}{x}}\) \(= \sqrt{\frac{45(100)}{45}}\) \(= \sqrt{100}\) \(= 10\) However, notice that I had to assume that 5x = 5x Why is this permissible in this problem? Even roots from negative numbers are undefined for the GMAT. We have \(\sqrt{5x}\) in one of the expressions, which means that x is nonnegative, thus 5x=5x. Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions
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11 Jun 2014, 01:46
5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2 How come you arrive to 25 from y? Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E.
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Re: NEW!!! Tough and tricky exponents and roots questions
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11 Jun 2014, 01:59
arindamsur wrote: 5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2 How come you arrive to 25 from y? Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. \(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2=25^2\) > . P.S. Please go this post: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628 (Writing Mathematical Formulas on the Forum). Thank you.
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Re: NEW!!! Tough and tricky exponents and roots questions
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11 Jun 2014, 02:09
\((2^{\sqrt{y}})^2=25^2\) Bunuel explain this part please Bunuel wrote: arindamsur wrote: 5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2 How come you arrive to 25 from y? Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. \(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2=25^2\) > . P.S. Please go this post: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628 (Writing Mathematical Formulas on the Forum). Thank you.
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Re: NEW!!! Tough and tricky exponents and roots questions
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11 Jun 2014, 02:11
arindamsur wrote: \((2^{\sqrt{y}})^2=25^2\) Bunuel explain this part please Bunuel wrote: Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\)? A. 14/5 B. 5 C. 28/5 D. 13 E. 14
First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. \(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2=25^2\) > . P.S. Please go this post: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628 (Writing Mathematical Formulas on the Forum). Thank you. We are given in the stem that \(2^{\sqrt{y}}=25\), thus \((2^{\sqrt{y}})^2=25^2\).
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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