Without-using-tables-or-calculator-compare-6-7-and-7-6-

Question Number 105301 by Don08q last updated on 27/Jul/20

Without using tables or calculator,

c o m p a r e 6^{7} and 7^{6}

Answered by 1549442205PVT last updated on 28/Jul/20

\frac{6^{7}}{7^{6}} = 6 \times {(\frac{6}{7})}^{6} > 6 \times {(\frac{4}{5})}^{6} = 6 \times {(\frac{64}{125})}^{2}

> 6 \times {(\frac{64}{128})}^{2} = 6 \times {(\frac{1}{2})}^{2} = \frac{6}{4} > 1

\Rightarrow 6^{7} > 7^{6}

Commented by Don08q last updated on 28/Jul/20

T h a n k y o u S i r!

Answered by JDamian last updated on 27/Jul/20

7^{6} = {(6 + 1)}^{6} =

= 6^{6} + 6 \cdot 6^{5} + 15 \cdot 6^{4} + 20 \cdot 6^{3} + 15 \cdot 6^{2} + 6 \cdot 6^{1} + 1 =

= 2 \cdot 6^{6} + 15 \cdot 6^{4} + 20 \cdot 6^{3} + 15 \cdot 6^{2} + 6 \cdot 6^{1} + 1 =

= 2 \cdot 6^{6} + 15 \cdot 6^{4} + 20 \cdot 6^{3} + 16 \cdot 6^{2} + 1

I f w e d i v i d e b o t h q u a n t i t i e s u n d e r t e s t b y 6^{6} :

\frac{6^{7}}{6^{6}} = 6

\frac{7^{6}}{6^{6}} = 2 + \frac{15}{6^{2}} + \frac{20}{6^{3}} + \frac{16}{6^{4}} + \frac{1}{6^{6}}

6 > 2 + \frac{15}{6^{2}} + \frac{20}{6^{3}} + \frac{16}{6^{4}} + \frac{1}{6^{6}}

4 > \frac{15}{6^{2}} + \frac{20}{6^{3}} + \frac{16}{6^{4}} + \frac{1}{6^{6}}

T h e r e f o r e 6^{7} > 7^{6}

Commented by Don08q last updated on 27/Jul/20

E x c e l l e n t . T h a n k y o u S i r .

Commented by malwaan last updated on 27/Jul/20

\frac{15}{6^{2}} < \frac{1}{2} < 1; \frac{20}{6^{3}} < \frac{1}{10} < 1

\frac{16}{6^{4}} = \frac{1}{81} ≪ 1; \frac{1}{6^{6}} ≪ 1

t h a n k y o u s i r J D a m i a n

Answered by mathmax by abdo last updated on 27/Jul/20

for x and y naturels and x > y let compare x^{y} and y^{x}

\Rightarrow let compare \ln (x^{y}) and \ln (y^{x}) (\ln is increazing function . .)

let φ (x) = ylnx - xlny for y fixed we have y ⩾ 1 and x \in] y, + \infty [

φ (y) = 0 and \lim_{x \to + \infty} φ (x) = \lim_{x \to + \infty} x (y \frac{lnx}{x} - lny) = - \infty

φ^{^{'}} (x) = \frac{y}{x} - lny \Rightarrow φ^{^{″}} (x) = - \frac{y}{x^{2}} < 0 \Rightarrow φ^{^{'}} is decreazing and

φ^{^{'}} (y) = 1 - lny < 0 \Rightarrow φ is decreazing \Rightarrow φ (x) < 0 \Rightarrow

ylnx - xlny < 0 \Rightarrow ylnx < xlny \Rightarrow x^{y} < y^{x}

x = 7 and y = 6 x > y \Rightarrow x^{y} < y^{x} \Rightarrow 7^{6} < 6^{7}

Commented by Don08q last updated on 28/Jul/20

T h a n k y o u S i r!

Commented by abdomsup last updated on 28/Jul/20

y o u a r e w e l c o m e