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

Question

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: $$(x+5)^{2n} + 50 = (x+20)^n+100$$, where $n =$ number of years (assuming $n\ge1$)

Then tried taking the log of both sides, but end up with: $$2n\log(x+5) = \log[(x+5)^n + 50]$$

How would I proceed from here, or must it be solved empirically?

Thanks in advance.

Answer 1

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't be raising to the power $n$. The first option gives you $45+5n$ in the $n^{\text{th}}$ half-year. After $m$ years, you have received $\sum_{i=1}^{2m}45+5i$ dollars in bonus. The second gives you $80+20m$ in the $m^{\text{th}}$ year. After $m$ years you have received $\sum_{i=1}^m80+20i$ dollars in bonus. Sum those and you have an analytic solution.

Answer 2

I'm not sure if taking logs will get you anywhere, have you tried writing $$(x+5)^{2n}=5^{2n}\left(1+\frac{x}{5}\right)^{2n}=5^{2n}\left(1+2n\frac{x}{5}+n(2n-1)\frac{x^2}{25}+.....\right)$$ and then the same for the $$(x+20)^{n}$$ and comparing the infinite sums as $n$ gets large.