Which is greater 4^300 or 3^400 ?

Let's write

4³⁰⁰       ???      3⁴⁰⁰

(4³)¹⁰⁰  ???   (3⁴)¹⁰⁰

64¹⁰⁰    ???       81¹⁰⁰

And it's obvious that the thing on the right is greater than the thing on the left.

So.....

4³⁰⁰       <      3⁴⁰⁰

And there you go....

Definition:

$$x = log_b(a) \Leftrightarrow b^x = a \,\,\,\, (for \,\, a,b >0 \,\, and \,\, b<>1)$$

Formula to convert logarithm base g to base b:

$$log_b(x) = log_b(g) * log_g( x )$$

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4^300 = number "a"  is a really large number, but we don't care what the actual number is, instead we want to know how many digits the number has:

4^300 = a 

 300 = log4( a  )

and:

 log10( a) = log10( 4 ) * log4( a ) = log10( 4 ) * 300

 log10( a) = log10( 4 ) * 300

log10( a ) = 180.61799739838872 ( so about 180 digits)

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3^400 = b

 400 = log3( b  )

and:

 log10( b) = log10( 3 ) * log3( b ) = log10( 3 ) * 400

 log10( b) = log10( 3 ) * 400

log10( b ) = 190.848501887864976 ( so about 190 digits)

Hi Chris,

That answer is really elegant.  I'm impressed!

Andre,

I have tried to work through yours - just in my head (which doesn't work very well)

But I don't get it.  Maybe I'll look again later.

Thanks....I actually didn't really know WHAT to do, at first. I was going to try something "complicated," but then I remembered a problem from a math "puzzles" book that I have that was similar to this. In fact, now that I think of it, I might post it to the forum just so some people can take a stab at it !!

That is an excellent idea.  You want to write or reference it somewhere in the puzzle threads.

 Reinout takes good care of them, and i really appreciate that,  but they could do with being organised a little better.  Maybe we should brain storm this one.  A lot of things need to be reorganised now that the forum has been revamped.

Hi Andre,

I have now looked at your answer properly.  It is really cool!

It is really great that one little  question got 2 totally different answers but both answers are really interesting and appropriate.