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Log problem (or by whatever method)

  • Thread starter tony24810
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  • #1
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Homework Statement



a^2 + 2^a = 100, where a is an integer, find a.

Homework Equations



all laws of indices and laws of log, I think

The Attempt at a Solution



By trial and error, answer can be easily determined, which is 6.

However, I am unsure how to approach this problem with algebraic approach. My attempts with log basically fail, because i ended up with log (a+b), which I don't know how to carry on.
 

Answers and Replies

  • #2
eumyang
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I don't think this can be solved algebraically (at least, with "elementary" algebra). I found a second solution graphically (graph y = x2 + 2x and y = 100 on a graphing calculator and find where the two graphs intersect). The second solution is not an integer.
 
  • #4
Ray Vickson
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Homework Statement



a^2 + 2^a = 100, where a is an integer, find a.

Homework Equations



all laws of indices and laws of log, I think

The Attempt at a Solution



By trial and error, answer can be easily determined, which is 6.

However, I am unsure how to approach this problem with algebraic approach. My attempts with log basically fail, because i ended up with log (a+b), which I don't know how to carry on.
Logs will not help. If 'a' is an integer, it must be a positive integer (can you see why?), so there are only a few possibilities, and you can easily try them out.

For positive integer a, the largest number of the form 2^a that is less than 100 is 64 = 2^6. You can test whether a = 6 solves the problem. If not, try a = 5, then a = 4, etc. Note that this method is not really 'trial and error'; it uses logic to cut the possibilities down to a small number.
 
  • #5
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Logs will not help. If 'a' is an integer, it must be a positive integer (can you see why?), so there are only a few possibilities, and you can easily try them out.

For positive integer a, the largest number of the form 2^a that is less than 100 is 64 = 2^6. You can test whether a = 6 solves the problem. If not, try a = 5, then a = 4, etc. Note that this method is not really 'trial and error'; it uses logic to cut the possibilities down to a small number.


If 'a' is an integer, it must be a positive integer (can you see why?)

As for your question, I figured that 2^(-ve integer) gives decimals, so no (-ve integer)^2 would always give integer, thus they never add up to 100, is that what you mean?
 

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