anonymous
  • anonymous
49log2*2log7 =? please do a favour to solve this ques.
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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rajat97
  • rajat97
it is in the form xlogy*ylogx
rajat97
  • rajat97
what can you do to simplify it?
Loser66
  • Loser66
.

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More answers

anonymous
  • anonymous
|dw:1436709166375:dw|
anonymous
  • anonymous
can i do somthing like that
Loser66
  • Loser66
I don't think you can do anything with the original one unless using calculator. It is simplest form since (2,7) are relative prime.
anonymous
  • anonymous
oke
anonymous
  • anonymous
49log2*2log7=7^2log2*2log7
anonymous
  • anonymous
what wud be the next step.
SolomonZelman
  • SolomonZelman
you are multiplying logarithms (correct?) \(\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) }\) At first I will ask you, are you sure that it is supposed to be a "•", and not a "+" ?
anonymous
  • anonymous
yes it is sure http://www.m4maths.com/placement-puzzles.php?SOURCE=amcat&TOPIC=Numerical%20Ability&SUB_TOPIC=Ratio%20and%20Proportion you can check out at this link
SolomonZelman
  • SolomonZelman
I don't have a login to this thing, but I of course believe you. I was just asking that for a clarification.....
anonymous
  • anonymous
oke but i wanna ask is there any formula to solve the above equation , which i hve done by drawing. log 2^7^2
SolomonZelman
  • SolomonZelman
I think that you can't make it any simpler than \(\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) =98\log(2)\log(7)}\) (And of course the assumed base is 10, not e)
anonymous
  • anonymous
yes!
SolomonZelman
  • SolomonZelman
Assumed base is also not important tho' for this multiplication, because the assumed base wouldn't matter with this multiplication.
SolomonZelman
  • SolomonZelman
All we did, is kinda: \(\large\color{black}{ \displaystyle 3a\cdot 4x=12ax }\)
SolomonZelman
  • SolomonZelman
we multiplied the coefficients, but you can't really put the logarithms together more than "log(2)log(7)"
anonymous
  • anonymous
yaa
SolomonZelman
  • SolomonZelman
questions about this?
anonymous
  • anonymous
so we need to put the values of log 2 and log 7 to calculate the exact answer?
anonymous
  • anonymous
as we cannt do anthing with them further
SolomonZelman
  • SolomonZelman
If you calculate these values it would be an approximation, and the "exact answer" is what we wrote before 98log(2)log(7)
anonymous
  • anonymous
49log2*2log7=7^2log2*2log7=7log2^2*2log7=7*2log7=2 this is the solution by somebody on another site . pls see to it
SolomonZelman
  • SolomonZelman
No, it is not equivalent to 2. THAT IS WRONG....
anonymous
  • anonymous
i didnt get the logic here
anonymous
  • anonymous
yaa
SolomonZelman
  • SolomonZelman
no logic... it is wrong... if it had logic it would be much more frightening:O
anonymous
  • anonymous
i have solved it , we have to take 2 as a base in frst case
SolomonZelman
  • SolomonZelman
I can't really follow what you did....
anonymous
  • anonymous
i have dun blunder to this ques *O*
anonymous
  • anonymous
hahaha
SolomonZelman
  • SolomonZelman
we can't take 2 as the base if it is a log.... you can't rewrite what's in the log, as the base... I mean you can't, but nothing correct is going to get out of it.
SolomonZelman
  • SolomonZelman
As again, the only thing here that I see valid is: \(\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) =98\log(2)\log(7)}\)
anonymous
  • anonymous
ok

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