Comparison between two power of numbers

[anemone]

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

 

[kaliprasad]

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

Spoiler

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

 

[anemone]

Spoiler

$8^{91} = 7^{91} * ( 1 + \frac{1}{7})^{91}$
$> 7^{91} * ( 1 + \frac{91}{7})$ 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)

 

[Albert1]

Spoiler

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

Spoiler

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 :

Spoiler

$8^{91}=(7+1)^{91}>7^{91}+91(7^{90})=7^{90}(7+91)=7^{90}(7\times7\times 2)=2\times 7^{92}>7^{92}$

 

[anemone]

Good job, Albert! (Cool)