Logarithm problem

debjit625 · 2013-02-10 03:53:39

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

Thanks

Last edited by debjit625 (2013-02-10 20:16:22)

anonimnystefy · 2013-02-10 03:57:42

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:
$gif.latex?\log_{10}{2}\cdot(2\log_{4}{5}+1)=1$

Do you know what $gif.latex?\frac{1}{\log_{a}{b}}$ is equal to?

debjit625 · 2013-02-10 04:08:23

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

anonimnystefy · 2013-02-10 04:21:43

Hi debjit625

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

debjit625 · 2013-02-10 04:33:14

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

Thanks

debjit625 · 2013-02-10 04:50:55

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

anonimnystefy · 2013-02-10 05:06:46

Hi

See the hidden text from my last post.

debjit625 · 2013-02-10 18:29:38

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

Thanks

Last edited by debjit625 (2013-02-10 20:14:15)

anonimnystefy · 2013-02-10 19:01:22

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

debjit625 · 2013-02-10 19:44:31

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

Bob · 2013-02-10 23:12:11

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 (2013-02-10 23:15:03)

debjit625 · 2013-02-10 23:54:16

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
$gif.latex?log_45=\frac{log_25}{log_24}=\frac{log_25}{2}$

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
$gif.latex?2\frac{log_25}{2}$

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

$gif.latex?2log_45%20+%201%20=%20log_210$

$gif.latex?2\frac{log_25}{2}%20+%201%20=%20log_2(2*5)$

$gif.latex?log_25%20+%201%20=%20log_22%20+%20log_25$

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

Last edited by debjit625 (2013-02-11 00:22:09)

anonimnystefy · 2013-02-10 23:55:39

Hi Roy

You need to remove all spaces from those links.

Bob · 2013-02-11 01:11:13

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

Math Is Fun Forum

#1 2013-02-10 03:53:39

Logarithm problem

#2 2013-02-10 03:57:42

Re: Logarithm problem

#3 2013-02-10 04:08:23

Re: Logarithm problem

#4 2013-02-10 04:21:43

Re: Logarithm problem

#5 2013-02-10 04:33:14

Re: Logarithm problem

#6 2013-02-10 04:50:55

Re: Logarithm problem

#7 2013-02-10 05:06:46

Re: Logarithm problem

#8 2013-02-10 18:29:38

Re: Logarithm problem

#9 2013-02-10 19:01:22

Re: Logarithm problem

#10 2013-02-10 19:44:31

Re: Logarithm problem

#11 2013-02-10 23:12:11

Re: Logarithm problem

#12 2013-02-10 23:54:16

Re: Logarithm problem

#13 2013-02-10 23:55:39

Re: Logarithm problem

#14 2013-02-11 01:11:13

Re: Logarithm problem

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