Prove that 2^{log_3(5)} equals 5^{log_3(2)}

ksdhart2 said:
That sure is a nice problem you have there... but what are your thoughts? What have you tried? Please re-read the Read Before Posting thread that's stickied at the top of each subforum, and comply with the rules laid out within. More specifically, please share with us any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.
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I tried using this blogbx = x but it gave me the same result.

​Any hints would be appreciated.
My thoughts are that it is more difficult than it seems and I don't know where to start.
 
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Roger.Robert said:
I tried using this blogbx = x but it gave me the same result.

​Any hints would be appreciated.
My thoughts are that it is more difficult than it seems and I don't know where to start.
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I'm not sure how you'd use that fact; can you show details of what you did?

One way to keep things from being too complicated is to just equate each side to a variable, x = 2^log3(5), y = 5^log3(2), and simplify log(x) and log(y). You should find that they are the same.

I'm sure there are many other ways to carry out the work. My first thought was to use the change of base formula.
 
Dr.Peterson said:
I'm not sure how you'd use that fact; can you show details of what you did?

One way to keep things from being too complicated is to just equate each side to a variable, x = 2^log3(5), y = 5^log3(2), and simplify log(x) and log(y). You should find that they are the same.

I'm sure there are many other ways to carry out the work. My first thought was to use the change of base formula.
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I tried to write 2 as \(\displaystyle \displaystyle\Large \frac{3}{\frac{3}{2}}\) ==> \(\displaystyle \displaystyle\Large (\frac{3}{\frac{3}{2}})^{\log _{3} 5}\) ==> \(\displaystyle \displaystyle\Large \frac{5}{\frac{5}{2^{\log _{3} 5}}}\) ==>
\(\displaystyle \displaystyle\Large 2^{log _{3} 5}\)

The change of base formula didn't work for me, got the same result again.
 
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Roger.Robert said:
I tried to write 2 as \(\displaystyle \displaystyle\Large \frac{3}{\frac{3}{2}}\) ==> \(\displaystyle \displaystyle\Large (\frac{3}{\frac{3}{2}})^{\log _{3} 5}\) ==> \(\displaystyle \displaystyle\Large \frac{5}{\frac{5}{2^{\log _{3} 5}}}\) ==>
\(\displaystyle \displaystyle\Large 2^{log _{3} 5}\)

The change of base formula didn't work for me, got the same result again.
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OK, that's creative, but you're right, it was a dead end.

If you want to directly go from the LHS to the RHS, one way to go would be to replace 2 on the left with 3^log3(2), and see what you can do with it.

This is effectively the same idea as mine and Halls', but carried out differently. Actually, I think I like this way more, though.

(I never actually pursued the change of base idea.)
 
My solution was similar, but it did involve the change-of-base formula at one point. Here were my steps:


  • \(\displaystyle 2^{log_3(5)} = 5^{log_3(2)}\)
  • \(\displaystyle log_3(5) = log_2 \left(5^{log_3(2)} \right)\) (Take log base 2 of both sides)
  • \(\displaystyle log_3(5) = log_3(2) \cdot log_2(5)\) (Power rule of logarithms)
  • \(\displaystyle \dfrac{log_3(5)}{log_2(5)} = log_3(2)\) (Divide both sides by \(\displaystyle log_2(5)\))
  • \(\displaystyle \dfrac{ln(2)}{ln(3)} = log_3(2)\) (Property of dividing logarithms)
  • \(\displaystyle log_3(2) = log_3(2)\) (Change-of-base formula)
 
As HallsofIvy already stated, taking log base 3 of both sides gives the desired results immediately.

log3(2log3(5)) = log3(5log3(2)) ==>log3(5)*log3(2)= log3(2)*log3(5).

This is really the cleanest way of doing such a problem.
 
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Jomo said:
As HallsofIvy already stated, taking log base 3 of both sides gives the desired results immediately.

log3(2log3(5)) = log3(5log3(2)) ==>log3(5)*log3(2)= log3(2)*log3(5).

This is really the cleanest way of doing such a problem.
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Perhaps; but one doesn't usually discover the "best" way immediately, so we need more prosaic ways to approach a problem that don't depend on a flash of insight. My first suggestion was intended to help in that process: anything you can do to get control of the complexity can help get the process started. (And it would lead to the very same result.) One thing I like to do is just "take a log" without worrying yet about what base to use (especially in a case like this with at least three bases to choose from). And it turns out that you'll get there whichever base you use.

One can also object that it is not so "clean" to show that the equation to be proved implies an identity, which is not the direction you really want to go. My second suggestion does it very cleanly, in a simple chain of equalities (not all of which really need to be written):

2log3(5) = (3log3(2))log3(5) = 3log3(2)log3(5) = 3log3(5)log3(2) =(3log3(5))log3(2) = 5log3(2).

But this, too, required the insight that using 3 as the base would be useful. I might have done this first, if my focus weren't on helping a student learn to solve problems on his own.

The most helpful thing, I think, is to explore different possibilities, as we've been doing.
 
Dr.Peterson said:
Perhaps; but one doesn't usually discover the "best" way immediately, so we need more prosaic ways to approach a problem that don't depend on a flash of insight. My first suggestion was intended to help in that process: anything you can do to get control of the complexity can help get the process started. (And it would lead to the very same result.) One thing I like to do is just "take a log" without worrying yet about what base to use (especially in a case like this with at least three bases to choose from). And it turns out that you'll get there whichever base you use.

One can also object that it is not so "clean" to show that the equation to be proved implies an identity, which is not the direction you really want to go. My second suggestion does it very cleanly, in a simple chain of equalities (not all of which really need to be written):

2log3(5) = (3log3(2))log3(5) = 3log3(2)log3(5) = 3log3(5)log3(2) =(3log3(5))log3(2) = 5log3(2).

But this, too, required the insight that using 3 as the base would be useful. I might have done this first, if my focus weren't on helping a student learn to solve problems on his own.

The most helpful thing, I think, is to explore different possibilities, as we've been doing.
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I see your point.
 
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