how to find the Greatest Number Among $3^{50} ,4^{40} ,5^{30}$ and $6^{20}$ please give a short cut method
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4What have you tried? Where do you get stuck? Especially what methods are you supposed to/want to use? – Ove Ahlman Jul 08 '16 at 11:20
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12If you start by taking the $10$th root of each of them, then paper and pencil will suffice. – hmakholm left over Monica Jul 08 '16 at 11:22
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For comparing $3^{50}$ and $5^{30}$, see http://math.stackexchange.com/a/1824321/589. – lhf Jul 08 '16 at 11:26
4 Answers
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Taking the tenth roots, you need to compare $3^5, 4^4, 5^3, 6^2$ since if $a > b$, $a^{10} > b^{10}$. These are $243, 256, 125, 36$. Thus the greatest is $4^{40}$
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Back in the older days, when we didn't have super powered calculators, such numbers were general dealt with by taking the log of them.
$$\log(3^{50})=50\log(3)$$
$$\log(4^{40})=40\log(4)$$
$$\log(5^{30})=30\log(5)$$
$$\log(6^{20})=20\log(6)$$
This should be much easier to put into a calculator and compare anyways.
Simply Beautiful Art
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You can simply take logarithm, or make exponents same:
$$3^{50}=\left(3^{5}\right)^{{10}};$$
$$4^{40}=\left(4^{4}\right)^{{10}}.$$
Davide Giraudo
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MrTanorus
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$\color\red{4^{40}}=2^{80}=256^{10}>243^{10}=$
$\color\red{3^{50}}=9^{25}>8^{25}=2^{75}>2^{70}=128^{10}>125^{10}=$
$\color\red{5^{30}}>4^{30}=2^{60}=8^{20}>$
$\color\red{6^{20}}$
barak manos
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