Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

bob bundy
2013-02-12 00:11:13

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

anonimnystefy
2013-02-11 22:55:39

Hi Roy

You need to remove all spaces from those links.

debjit625
2013-02-11 22:54:16

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
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...

bob bundy
2013-02-11 22:12:11

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

debjit625
2013-02-11 18:44:31

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

anonimnystefy
2013-02-11 18:01:22

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

debjit625
2013-02-11 17:29:38

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

anonimnystefy
2013-02-11 04:06:46

Hi

See the hidden text from my last post.

debjit625
2013-02-11 03:50:55

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

debjit625
2013-02-11 03:33:14

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

Thanks

anonimnystefy
2013-02-11 03:21:43

Hi debjit625

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

debjit625
2013-02-11 03:08:23

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
2013-02-11 02:57:42

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:
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?

debjit625
2013-02-11 02:53:39

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

Board footer

Powered by FluxBB