Prove that $5^{44}<4^{53}$
I tried to prove that $5^{44}<4^{52}$
$\implies 5^{11}<4^{13}$
But couldn't
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1$\begingroup$ The two numbers aren't particularly close...as a hint, use $2^7>5^3$. $\endgroup$– luluCommented Dec 26, 2020 at 14:43
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$\begingroup$ Thanks now I got it. How did you find the solution. $\endgroup$– saket kumarCommented Dec 26, 2020 at 14:52
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$\begingroup$ Like how you approached it, $\endgroup$– saket kumarCommented Dec 26, 2020 at 14:52
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$\begingroup$ Since $2^7$ is just a little greater than $2^5$ I figured it was likely that we could use those numbers as a base. Can't really be sure till you write it out, of course. Worth noting that the same calculation actually shows that $2^{103}>5^{44}$. $\endgroup$– luluCommented Dec 26, 2020 at 15:06
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1$\begingroup$ This is an abstract duplicate of Without using a calculator and logarithm, which of $100^{101} , 101^{100}$ is greater?. Same problem, different numbers. Found using Approach0. $\endgroup$– Toby MakCommented Dec 27, 2020 at 7:52
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1 Answer
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Note that $ 4^{53} = 2^{106} , $ so you need to show that $$2^{106} > 5^{44} \; . $$
Now try to find a simpler inequality of that form $2^x > 5^y$: $$ 2^7 = 128 > 125 = 5^3 $$ Now we can use this to prove the original inequality: $$ 2^{106} = 2 \cdot 2^{105} = 2 \cdot \left( 2^7 \right)^{15} > 2 \cdot \left( 5^3 \right)^{15} = 2 \cdot 5^{45} > 5^{45} > 5^{44} \; . $$
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1$\begingroup$ Thanks bro, was looking for this $\endgroup$ Commented Dec 26, 2020 at 14:55
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1$\begingroup$ Still a typo in that ( ), you want $5^3$, not $2^3$. $\endgroup$ Commented Dec 26, 2020 at 14:55