Can ( x + 5 ) 2 n + 50 = ( x + 20 ) n + 100 be simplified algebraically to give a solution?I have a brainteaser book which I'm working through and going overboard on. One puzzle is as follows: Your boss offers you a choice of two bonuses (1) fifty dollars after six months and an ongoing semiannual increase of five dollars; or (2) one hundred dollars after a year and an ongoing annual increase of twenty dollars. Which bonus will prove more lucrative? Rather than writing out a table, I wanted to calculate at what time one or the other becomes more lucrative.I wrote out: ( x + 5 ) 2 n + 50 = ( x + 20 ) n + 100, where n = number of years (assuming n ≥ 1)Then tried taking the log of both sides, but end up with: 2 n log ⁡ ( x + 5 ) = log ⁡ [ ( x + 5 ) n + 50 ]How would I proceed from here, or must it be solved empirically?Thanks in advance.

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Can   (  x  +  5      )          2      n        +  50  =  (  x  +  20      )    n    +  100 be simplified algebraically to give a solution?I have a brainteaser book which I&#039;m working through and going overboard on. One puzzle is as follows: Your boss offers you a choice of two bonuses (1) fifty dollars after six months and an ongoing semiannual increase of five dollars; or (2) one hundred dollars after a year and an ongoing annual increase of twenty dollars. Which bonus will prove more lucrative? Rather than writing out a table, I wanted to calculate at what time one or the other becomes more lucrative.I wrote out:  (  x  +  5      )          2      n        +  50  =  (  x  +  20      )    n    +  100, where   n  = number of years (assuming   n  ≥  1)Then tried taking the log of both sides, but end up with:  2  n  log  ⁡  (  x  +  5  )  =  log  ⁡  [  (  x  +  5      )    n    +  50  ]How would I proceed from here, or must it be solved empirically?Thanks in advance.

Prezrenjes0n · Accepted Answer

You have not written the correct equation. You are using   x for your current salary, but you can ignore that. The increases are additive, so you shouldn&#039;t be raising to the power   n. The first option gives you   45  +  5  n in the       n          th       half-year. After   m years, you have received       ∑          i      =      1              2      m        45  +  5  i dollars in bonus. The second gives you   80  +  20  m in the       m          th       year. After   m years you have received       ∑          i      =      1        m    80  +  20  i dollars in bonus. Sum those and you have an analytic solution.

Jorge Schmitt · Accepted Answer

I&#039;m not sure if taking logs will get you anywhere, have you tried writing  (  x  +  5      )          2      n        =      5          2      n                  (      1      +              x        5            )              2      n        =      5          2      n            (    1    +    2    n          x      5        +    n    (    2    n    −    1    )                  x        2            25        +    .    .    .    .    .    )  and then the same for the  (  x  +  20      )          n      and comparing the infinite sums as   n gets large.

Can (x+5)^(2n)+50=(x+20)^(n)+100 be simplified algebraically to give a solution?

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