Proving $3^{100} > 5\cdot10^{47}$ with integral representation

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I want to prove that $3^{100} > 5\cdot10^{47}$ without using calculator or any approximation of the logarithm. I tought about finding an integral representation of $(3^{100} - 5\cdot10^{47})$ with the integrand non negative in the chosen integration interval (something like this), but I don't know how to chose the integrand. Do you have any ideas? Or even any elegant way to prove the inequality?

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We have by hand calculation

  • $3^{10} =59049>59\cdot 10^3$
  • $59^2=3481>348\cdot10$
  • $348^2=121104>12\cdot 10^4$

then

$$3^{100}>348\cdot 12^2\cdot 10^{43}>5.01\cdot 10^{47}$$