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anonymous
 3 years ago
log
anonymous
 3 years ago
log

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Log of nothing is nothing .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[If \log 5 x=y, then 5^{5y}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you sure that '5' is adjacent to that 'x', and not the base of log?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.15^y * 5^y * 5^y * 5^y * 5^y = 5^{5y} x^5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was not able to type it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Hero @saifoo.khan @.Sam. @dpaInc @jim_thompson5910 @lgbasallote @Limitless @UnkleRhaukus

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\log(0)\) is actually undefined, @apoorvk.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sure then: you gotta use this theorem: \[\large a^{log_bc} = c^{log_b a}\] dw:1339384039980:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah @Limitless  i was joking up there^ :p

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1Yuppie i was correct...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes I was choosing you as best answer but apoorvk gave steps

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Create a new thread for a new question @ajaykharabe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/4fd561fee4b04bec7f1698ee

mathslover
 3 years ago
Best 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mathslover  you deserve this medal, you posted the solution first :) And no, am not an expert  far from it!! :)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1apoorvk 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 :)
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