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Joined: 02 Sep 2009
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61% (01:35) correct 39% (02:14) wrong based on 120 sessions
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Math Expert
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Re M0716
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19 Jan 2015, 04:13
Had problems with the website (became irresponsive) during this question causing me to spend almost 12 mins in it. Couldn't finish the test after.



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19 Jan 2015, 04:14



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26 Jun 2015, 15:20
I'm a little rusty, how do you cross multiply to get everything to be a divisor of 1 to being the number alone to the exponent. That is where I struggled with this problem. Thanks.



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27 Jun 2015, 03:43



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Re: M0716
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22 Dec 2015, 07:16
Not sure I am following when you equate the exponents of 7. Would that not be 18y=x?
Thanks for the help.



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22 Dec 2015, 07:22



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Re: M0716
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17 Jul 2016, 10:43
Hello,
How does the left hand side of this equation factor? ie the 28^18y
Sorry, a bit rusty on the the exponent rules. Thanks in advance.



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17 Jul 2016, 10:49



Math Expert
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17 Jul 2016, 10:49



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Re: M0716
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07 Aug 2016, 15:40
I rewrote the equation as: 7^x∗8^12=28^18y 7^x∗2^3(12) = 7^18y∗ 2^2(18y) 7^x∗2^36 = 7^18y∗ 2^36y from here we can see that x = 18y or x = 18y and 36=36y or y =1. If y is 1 then x = 18 yx = 118 = 17



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24 Jun 2017, 09:52
BunuelHi Bunuel, if x & y are not integers, does it give infinitely many solutions of x & y, or does it still give finite (but different) solutions? Thanks!



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24 Jun 2017, 10:43



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01 Aug 2018, 01:19
i did this question in following way: simplified the given equation into 7^(18yx) = 2^36(1y) now as 7 and 2 are prime no.s , LHS will be equal to RHS only when when respective powers of 7 and 2 is 0 hence, i put 18yx=0, this implies 18y=x. and 36(1y)=0, this implies y=1 hence yx= 118= 17
is above method correct?



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01 Aug 2018, 01:24










