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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- rajat97

it is in the form
xlogy*ylogx

- rajat97

what can you do to simplify it?

- Loser66

.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

|dw:1436709166375:dw|

- anonymous

can i do somthing like that

- 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

oke

- anonymous

49log2*2log7=7^2log2*2log7

- anonymous

what wud be the next step.

- 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

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

I don't have a login to this thing, but I of course believe you. I was just asking that for a clarification.....

- 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

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

yes!

- SolomonZelman

Assumed base is also not important tho' for this multiplication, because the assumed base wouldn't matter with this multiplication.

- SolomonZelman

All we did, is kinda: \(\large\color{black}{ \displaystyle 3a\cdot 4x=12ax }\)

- SolomonZelman

we multiplied the coefficients, but you can't really put the logarithms together more than "log(2)log(7)"

- anonymous

yaa

- SolomonZelman

questions about this?

- anonymous

so we need to put the values of log 2 and log 7 to calculate the exact answer?

- anonymous

as we cannt do anthing with them further

- 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

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

No, it is not equivalent to 2. THAT IS WRONG....

- anonymous

i didnt get the logic here

- anonymous

yaa

- SolomonZelman

no logic... it is wrong... if it had logic it would be much more frightening:O

- anonymous

i have solved it , we have to take 2 as a base in frst case

- SolomonZelman

I can't really follow what you did....

- anonymous

i have dun blunder to this ques *O*

- anonymous

hahaha

- 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

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

ok

Looking for something else?

Not the answer you are looking for? Search for more explanations.