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Comment on Exponent Laws - Part II
https://gre.myprepclub.com/forum
How was 98^7/7^63 rewritten as 7^14*2^7/7^63
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Question link: https://gre.myprepclub.com/forum/which-is-greater-98-7-7-63-or-4501.html
I think you're referring to someone else's solution, but I'm happy to help.
First recognize that 98 = 7 x 7 x 2
So, 98^7 = (7 x 7 x 2)^7
= 7^7 x 7^7 x 2^7 [power of a product rule]
= (7^7 x 7^7) x 2^7
= (7^14) x 2^7
Does that help?
By the way, here's my full solution: https://gre.myprepclub.com/forum/which-is-greater-98-7-7-63-or-4501.html...
Cheers,
Brent
https://gre.myprepclub.com/forum
I tried to solve this question by inputting values. I used x = 2 and 5 and got B as the answer both times. But the solution is D. How can I avoid making these types of mistakes when both the input values give a particular answer?
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Question link: https://gre.myprepclub.com/forum/x-is-an-integer-greater-than-1724.html
Yes, this is the problem with testing values. Unless you get two conflicting results, you can never be certain of the correct answer.
Here's a rudimentary example of what I mean:
GIVEN: x is a prime number
QUANTITY A: x
QUANTITY B: 5
So, let's test some possible values of x
x COULD equal 7 (a prime), in which case Quantity A is greater.
x COULD equal 53 (a prime), in which case Quantity A is greater.
x COULD equal 19 (a prime), in which case Quantity A is greater.
x COULD equal 37 (a prime), in which case Quantity A is greater.
At this point, it APPEARS to be the case that the correct answer is A.
However, had we chosen better numbers (e.g., x = 3, x = 5 and x = 7), we would have seen that the correct answer is actually D.
In the question you've asked about, we have powers of 3 and 4. Since those powers get quite large very quickly, it's a good idea to start with smaller numbers and keep going. Also note that there will be times when you have to test 3 or 4 or 5 or even more numbers before you're able to be certain about the correct answer.
At 1:37 of the following video, I discuss some "nice" numbers to test: https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...
Noted. Do you think that when
Yes. In most cases, an
Yes. In most cases, an algebraic solution is faster than testing the answer choices.
Cheers,
Brent
81^3 + 27^4 is equivalent to
Indicate all such expressions.
I) (37)(2)
II) (312)(2)
III) (96)(2)
IV) 912
V) 324
Please help me out in this question.
https://gre.myprepclub.com/forum/qotd-1-81-3-27-4-is-equivalent-to-which-2302.html
Question link: https:/
Question link: https://gre.myprepclub.com/forum/qotd-1-81-3-27-4-is-equivalent-to-which...
Here's my solution: https://gre.myprepclub.com/forum/qotd-1-81-3-27-4-is-equivalent-to-which...
Cheers,
Brent
Regarding the 3rd
https://gre.myprepclub.com/forum/if-5-25-4-30-2-10-k-what-is-the-10611.html
5^2^5 * 4^1^3 = 2 * 10^k
, what is the value of K
At a certain point you say, "we can focus solely on the 5's". Further down on the page, someone else says: "Now you can equate either the exponent of 5 or 2."
I'm not familiar with this. So, if there are exponents with similar bases, I can just equate the exponents and ignore everything else in the equation? Is this a rule? Is there a name for it? Is this common knowledge? I guess it makes sense, I just don't remember ever being told about this.
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Question link: https://gre.myprepclub.com/forum/if-5-25-4-30-2-10-k-what-is-the-10611.html
If we know that all of the values are positive integers, then we can focus on just 1 base.
Here's an analogous question:
At a certain school, there are 2x + 7 girls, and there are 3x - 1 boys. If there are 17 girls and 14 boys, what is the value of x?
Notice that we can focus on EITHER the girls (i.e., 2x + 7 = 17)) OR the boys (i.e., 3x - 1 = 14) to find the value of x.
Similarly, if we have the equation: [3^(2x + 1)][7^(5x - 4)] = (3^5)(7^6), we can focus on either the powers with base 3 (i.e., 2x + 1 = 5) or the powers with base 7 (i.e., 3x - 1 = 14).
Of course, you'll still get the correct answer if you don't focus on just one set of powers. It will just take you a tiny bit longer.
Does that help?
Cheers,
Brent
The analogy is helpful, and
To make sure though, this rule works as long as there no other factors floating around or other things in the equation, correct? For example, if I have just: (2^26) = (2^k). Then obviously, k = 26. But what if I have (2^26) = 5(2^k). I have the same base with 2^26, and 2^k, but I have that extra 5^1 that doesn't have a corresponding base on the other side. I assume I can't say k = 26 in that case, or can I? What if it was 5 + 2^k = 2^26?
I solved the problem the long way ending with 10^25 = 10^k , but I like the idea of being able to use this technique. I just want to make sure I understand when I can use it.
As long as we know the
As long as we know the exponents are INTEGERS, we can focus on one of the bases. Here's why:
If we have 2^n = 35, there's a NON-INTEGER value of n (between 5 and 6) that satisfies the equation.
Likewise, there's a non-integer value of x (between 2 and 3) that satisfies the equation 7^x = 121
So, say for example, we have the equation (7^x)(11^y) = (7^3)(11^2)
If we allow x and y to have NON-INTEGER values, it COULD be the case that (11^y) = 7^3, and (7^x = 11^2)
Does that help?
Cheers,
Brent
Yes, that helps. It all
https://gre.myprepclub.com/forum
On that question how did you go from (0.3)^20 to 10^20?
Simplify both quantities to get:
Quantity A: 10^20
Quantity B: (0.03)^30
Thanks
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Question link: https://gre.myprepclub.com/forum/which-is-greater-0-3-20-or-17359.html
The property here is as follows: (x^n)/(y^n) = (x/y)^n
For example: (24^7)/(8^7) = (24/8)^7 = 3^7
Likewise: (0.3^20)/(0.03^20) = (0.3/0.03)^20 = 10^20
Does that help?