Comparing Powers: A Mental Technique

  • Thread starter Thread starter vin300
  • Start date Start date
  • Tags Tags
    Compare
vin300
Messages
602
Reaction score
4
This is by no means a trick question, I would want to know whether there is any technique to compare two powered terms with different bases, that is if I say I want to compare 2 powered 200 to 3 powered 120 to estimate which of these may be greater without any special or digital aid, how can I do it in my mind?
 
Mathematics news on Phys.org
Take the logarithm of both numbers:

ln(2200) = 200 ln 2 = 138.629436112
ln(3120) = 120 ln 3 = 131.83347464

So 2200 > 3120

EDIT: fixed a math error.
 
Are you talking about exponents?
So like x^n vs y^m?
You could just take the logs, and do some algebra.

Have the ~ symbol be a placeholder for =,<,> symbols

2^{200} \sim 3^{120}
200 log(2) \sim 120 log(3)
\frac{200}{120} \sim \frac{log(3)}{log(2)}
\frac{5}{3} \sim log_2(3)
2^{\frac{5}{3}} \sim 3
2^5 \sim 3^3
32 \sim 27

therefore: ~ is >
so
2^{200} &gt; 3^{120}

No calculator needed.
 
Thanks.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top