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apoorvkBest ResponseYou've already chosen the best response.4
Log of nothing is nothing .
 one year ago

ajaykharabeBest ResponseYou've already chosen the best response.0
\[If \log 5 x=y, then 5^{5y}\]
 one year ago

apoorvkBest ResponseYou've already chosen the best response.4
you sure that '5' is adjacent to that 'x', and not the base of log?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.1
5^y * 5^y * 5^y * 5^y * 5^y = 5^{5y} x^5
 one year ago

ajaykharabeBest ResponseYou've already chosen the best response.0
I was not able to type it
 one year ago

ajaykharabeBest ResponseYou've already chosen the best response.0
@Hero @saifoo.khan @.Sam. @dpaInc @jim_thompson5910 @lgbasallote @Limitless @UnkleRhaukus
 one year ago

LimitlessBest ResponseYou've already chosen the best response.0
\(\log(0)\) is actually undefined, @apoorvk.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.4
sure then: you gotta use this theorem: \[\large a^{log_bc} = c^{log_b a}\] dw:1339384039980:dw
 one year ago

apoorvkBest ResponseYou've already chosen the best response.4
yeah @Limitless  i was joking up there^ :p
 one year ago

mathsloverBest ResponseYou've already chosen the best response.1
Yuppie i was correct...
 one year ago

ajaykharabeBest ResponseYou've already chosen the best response.0
yes I was choosing you as best answer but apoorvk gave steps
 one year ago

apoorvkBest ResponseYou've already chosen the best response.4
Create a new thread for a new question @ajaykharabe
 one year ago

ajaykharabeBest ResponseYou've already chosen the best response.0
http://openstudy.com/study#/updates/4fd561fee4b04bec7f1698ee
 one year ago

mathsloverBest ResponseYou've already chosen the best response.1
@ajaykharabe i dont need medal ... apoorvk is one of the best experts here .. no one can beat him ..............................it is nice to see that u got it :) that is why there is openstudy
 one year ago

apoorvkBest ResponseYou've already chosen the best response.4
@mathslover  you deserve this medal, you posted the solution first :) And no, am not an expert  far from it!! :)
 one year ago

mathsloverBest ResponseYou've already chosen the best response.1
apoorvk i know that i answered it first but the one who explained is you ... no problem that u r not an expert but u r playing the role of expert :)
 one year ago
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