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#1 2013-02-11 02:53:39

debjit625
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Logarithm problem

May be its simple but I can't solve it...
Show that : log2(2log(base 4) 5 + 1) = 1
http://latex.codecogs.com/gif.latex?log_{10}2(2log_{4}5+1)%20=%201

Thanks

Last edited by debjit625 (2013-02-11 19:16:22)

Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

#2 2013-02-11 02:57:42

anonimnystefy
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Re: Logarithm problem

Hi debjit625

The LaTeX on this forum isn't functioning at the moment so here is the picture with what you presumably want to show:
http://latex.codecogs.com/gif.latex?\log_{10}{2}\cdot(2\log_{4}{5}+1)=1

Do you know what http://latex.codecogs.com/gif.latex?\frac{1}{\log_{a}{b}} is equal to?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón

#3 2013-02-11 03:08:23

debjit625
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Re: Logarithm problem

Ok I was having problem with  Latex.... and yes thats write.

1/log(base a)b = log(base b)a ,is that what you wanted to know

Thanks

Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

#4 2013-02-11 03:21:43

anonimnystefy
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Re: Logarithm problem

Hi debjit625

Yes, that is the one. Do you see how you can use it here?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón

#5 2013-02-11 03:33:14

debjit625
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Re: Logarithm problem

No,not sure.
I can use that inside the bracket to solve log(base 4) 2 to log/(base2)4 ....

Thanks

Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

#6 2013-02-11 03:50:55

debjit625
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Re: Logarithm problem

I am not sure how to do it ...still confused

Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

#7 2013-02-11 04:06:46

anonimnystefy
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Re: Logarithm problem

Hi

See the hidden text from my last post.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón

#8 2013-02-11 17:29:38

debjit625
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Re: Logarithm problem

Sorry I cant understand...
dividing both the sides by "log (base 10) 2" will give us
http://latex.codecogs.com/gif.latex?2log_45+1%20=%20log_210
2log(base 4)5 + 1 = log (base 2) 10

Thanks

Last edited by debjit625 (2013-02-11 19:14:15)

Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

#9 2013-02-11 18:01:22

anonimnystefy
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Re: Logarithm problem

That is correct. Can you proceed from here or do you need the next step?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón

#10 2013-02-11 18:44:31

debjit625
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Re: Logarithm problem

I need the next step ...
I am not sure how to prove LHS is equal to 1,shouldn't we only work with LHS and prove/show it is 1?

Thanks

Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

#11 2013-02-11 22:12:11

bob bundy
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Re: Logarithm problem

hi debjit625

I'd get everything in the same log base.  As it is easy to get log base 4 into log base 2 that's the next step:

log(base4)5 = log(base2)5/log(base2)4 = (log(base2)5)/2

so your expression (from post 8) becomes (all logs now in base 2):

(2log5)/2 + 1 = (2log5)/2 + log2

Should be easy to finish from there.

Bob

Last edited by bob bundy (2013-02-11 22:15:03)

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#12 2013-02-11 22:54:16

debjit625
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Re: Logarithm problem

Well that solved the problem ,but still I have questions...
I understood that you used change base formula on LHS to change the base of
http://latex.codecogs.com/gif.latex?log_45=\frac{log_25}{log_24}=\frac{log_25}{2}

bob bundy wrote:

hi debjit625
so your expression (from post 8) becomes (all logs now in base 2):
(2log5)/2 + 1 = (2log5)/2 + log2
Bob

But what I didnt understood is that how you got it on RHS
http://latex.codecogs.com/gif.latex?2\frac{log_25}{2}

As per me its like this
http://latex.codecogs.com/gif.latex?log_{10}2(2log_45+1)%20=%201

http://latex.codecogs.com/gif.latex?2log_45%20+%201%20=%20log_210

http://latex.codecogs.com/gif.latex?2\frac{log_25}{2}%20+%201%20=%20log_2(2*5)

http://latex.codecogs.com/gif.latex?log_25%20+%201%20=%20log_22%20+%20log_25

Thanks everybody it seems I have to learn a lot ,off course from you guys...

Last edited by debjit625 (2013-02-11 23:22:09)

Debjit Roy
___________________________________________________
The essence of mathematics lies in its freedom - Georg Cantor

#13 2013-02-11 22:55:39

anonimnystefy
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Re: Logarithm problem

Hi Roy

You need to remove all spaces from those links.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón

#14 2013-02-12 00:11:13

bob bundy
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Re: Logarithm problem

hi debjit625

What you have done is equivalent to my version.

Strictly, you should have LHS = ... = ... = ... = RHS.

But you have all the elements to re-write it like that now.

Hints:  1 = log(base n) n      for all n                    and       log(base a)b x log(base b) a = 1 for all a and b

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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