if x1 and x2 are the roots of 3^x-2 + 1= -3^4-x then x1+x2=?

Question

ganeshie8 · Answer

$ \large   3^{x-2} + 1= -3^{4-x}$

ganeshie8 · Answer

like that ?

anonymous · Answer

yes @ganeshie8

ganeshie8 · Answer

okay isolate the exponent term first

ganeshie8 · Answer

use below :
$\large a^{m \pm  n} = a^m  .  a^{\pm n} $

ganeshie8 · Answer

$\large   3^{x-2} + 1= -3^{4-x}$

$\large   3^{x}.3^{-2} + 1= -3^{4}3^{-x}$

anonymous · Answer

oky then what should i do?

anonymous · Answer

will it be $$3^{x+x} = -3^{7-3}$$
$$3^{2x} = -3^{4}$$

anonymous · Answer

oky!!!

ganeshie8 · Answer

$\large   3^{x-2} + 1= -3^{4-x}$

multiply thru by 3^2 :
$\large   3^{x} + 3^2 = -3^{6-x}$

multiply again by 3^x :
$\large   \left(3^{x}\right)^2 + 3^2\left(3^x\right)= -3^{6}$

ganeshie8 · Answer

$\large   \left(3^{x}\right)^2 + 3^2\left(3^x\right)+ 3^{6}=0$

ganeshie8 · Answer

Notice that this is a quadratic in 3^x

product of roots  = c/a =  ?

anonymous · Answer

will it be $$\frac{ 3^{2} }{ (3x)^{2} }$$

ganeshie8 · Answer

nope, may be let 3^x  = t for the time being

ganeshie8 · Answer

then the equation becomes :

$\large t^2  + 3^2t + 3^6 = 0$

ganeshie8 · Answer

product of roots =  $\large   t_1 t_2 =  3^6$

ganeshie8 · Answer

$\large \implies      \rm 3^{x_1}3^{x_2} =  3^6$

ganeshie8 · Answer

$\large \implies      \rm 3^{x_1 + x_2} =  3^6$

ganeshie8 · Answer

$\large \rm \implies     x_1 + x_2 = 6$

ganeshie8 · Answer

see if that makes more or less sense..

anonymous · Answer

ohhh yaaa!!!