49log2*2log7 =? please do a favour to solve this ques.

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49log2*2log7 =? please do a favour to solve this ques.

Linear Algebra
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it is in the form xlogy*ylogx
what can you do to simplify it?
.

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|dw:1436709166375:dw|
can i do somthing like that
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.
oke
49log2*2log7=7^2log2*2log7
what wud be the next step.
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 "+" ?
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
I don't have a login to this thing, but I of course believe you. I was just asking that for a clarification.....
oke but i wanna ask is there any formula to solve the above equation , which i hve done by drawing. log 2^7^2
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)
yes!
Assumed base is also not important tho' for this multiplication, because the assumed base wouldn't matter with this multiplication.
All we did, is kinda: \(\large\color{black}{ \displaystyle 3a\cdot 4x=12ax }\)
we multiplied the coefficients, but you can't really put the logarithms together more than "log(2)log(7)"
yaa
questions about this?
so we need to put the values of log 2 and log 7 to calculate the exact answer?
as we cannt do anthing with them further
If you calculate these values it would be an approximation, and the "exact answer" is what we wrote before 98log(2)log(7)
49log2*2log7=7^2log2*2log7=7log2^2*2log7=7*2log7=2 this is the solution by somebody on another site . pls see to it
No, it is not equivalent to 2. THAT IS WRONG....
i didnt get the logic here
yaa
no logic... it is wrong... if it had logic it would be much more frightening:O
i have solved it , we have to take 2 as a base in frst case
I can't really follow what you did....
i have dun blunder to this ques *O*
hahaha
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.
As again, the only thing here that I see valid is: \(\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) =98\log(2)\log(7)}\)
ok

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