No. of real solution of exponential equation

[juantheron]

No. of real solution of the equation $1+8^x+27^x = 2^x+12^x+9^x.$

 

[Albert1]

No. of real solution of the equation $1+8^x+27^x = 2^x+12^x+9^x.$

Spoiler

$1+8^x+27^x =1+8^x+ (2+12+9+4)^x----(1)$
$=1+8^x+(2^x+12^x+9^x)+4^x+-----$
$=2^x+12^x+9^x----(2)$
if $x\neq 0$ then (1) > (2)
$\therefore x=0$ is the only solution

 

[MarkFL]

Spoiler

$1+8^x+27^x =1+8^x+ (2+12+9+4)^x----(1)$
$=1+8^x+(2^x+12^x+9^x)+4^x+-----$
$=2^x+12^x+9^x----(2)$
if $x\neq 0$ then (1) > (2)
$\therefore x=0$ is the only solution

Spoiler

You have employed a mistake known as "The Freshman's Dream"...:D

 

[Albert1]

Spoiler

You have employed a mistake known as "The Freshman's Dream"...:D

I did not say:$27^x=(2+9+12+4)^x=2^x+9^x+12^x+4^x$
I said :$27^x=2^x+9^x+12^x+4^x+----$
the remaining terms are omitted

 

[kaliprasad]

I did not say:$27^x=(2+9+12+4)^x=2^x+9^x+12^x+4^x$
I said :$27^x=2^x+9^x+12^x+4^x+----$
the remaining terms are omitted

Spoiler

wrong

to give an example $3^{.5} = 1.7 < 2^{.5} + 1^{.5 }$ and not > (given approximately)

 

[Albert1]

Spoiler

wrong

to give an example $3^{.5} = 1.7 < 2^{.5} + 1^{.5 }$ and not > (given approximately)

let $f(x)=1+8^x+27^x-----(1)$
$g(x)=2^x+9^x+12^x----_(2)$
if $x>0$ then (1)>(2)
if $x=0$ then (1)=(2)=3
if $x<0$ and $ x \rightarrow -\infty$
then $f(x)\rightarrow 1$,and $g(x)\rightarrow 0$
here $f(x)$ and $g(x)$ both are increasing on $(-\infty , \infty)$