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 need to remove all spaces from those links.

]]>I understood that you used change base formula on LHS to change the base of

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

As per me its like this

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

]]>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

]]>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

]]>dividing both the sides by "log (base 10) 2" will give us

2log(base 4)5 + 1 = log (base 2) 10

Thanks

]]>See the hidden text from my last post.

]]>I can use that inside the bracket to solve log(base 4) 2 to log/(base2)4 ....

Thanks

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

]]>

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

Thanks

]]>The LaTeX on this forum isn't functioning at the moment so here is the picture with what you presumably want to show:

Do you know what is equal to?

]]>Show that : log2(2log(base 4) 5 + 1) = 1

Thanks

]]>