MHB Comparison between two power of numbers

anemone
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Prove that $8^{91}\gt 7^{92}$.
 
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anemone said:
Prove that $8^{91}\gt 7^{92}$.

$8^{91} = 7^{91} * ( 1 + \frac{1}{7})^{91}$
$> 7^{91} * ( 1 + \frac{91}{7})$ using for n integer > 1 $(1+x)^n > 1 + nx$ as other terms are positive
$> 7^{91} * 14 > 2* 7^{92} > 7^{92}$
 
kaliprasad said:
$8^{91} = 7^{91} * ( 1 + \frac{1}{7})^{91}$
$> 7^{91} * ( 1 + \frac{91}{7})$ using for n integer > 1 $(1+x)^n > 1 + nx$ as other terms are positive
$> 7^{91} * 14 > 2* 7^{92} > 7^{92}$

Very well done, kaliprasad!(Cool)

There is another way to crack the problem, I welcome others to take a stab at it! (Sun)
 
kaliprasad said:
$8^{91} = 7^{91} * ( 1 + \frac{1}{7})^{91}$
$> 7^{91} * ( 1 + \frac{91}{7})$ using for n integer > 1 $(1+x)^n > 1 + nx$ as other terms are positive
$> 7^{91} * 14 > 2* 7^{92} > 7^{92}$
it should be:
$8^{91} = 7^{91} * ( 1 + \frac{1}{7})^{91}$
$> 7^{91} * ( 1 + \frac{91}{7})$ using for n integer > 1 $(1+x)^n > 1 + nx$ as other terms are positive
$= 7^{91} * 14 = 2* 7^{92} > 7^{92}$
my solution :using binomial expansion :
$8^{91}=(7+1)^{91}>7^{91}+91(7^{90})=7^{90}(7+91)=7^{90}(7\times7\times 2)=2\times 7^{92}>7^{92}$
 
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Good job, Albert! (Cool)
 
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