Question

ajudem.resolva essa equação exponencial $<var>3^{x-1}+3^{x+1}+3^{x+2}=3^{x}+306</var>$

Answer 1

$<var>3^x^-^1+3^x^+^1+3^x^+^2=3^x+306</var>$

 

$<var>3^x *3^-^1 + 3^x * 3^1 + 3^x*3^2=3^x+306</var>$

 

$<var>3^x = y</var>$

 

$<var> \frac{y}{3}+3y+9y-y=306\\ \frac{y}{3}+11y=306</var>$

 

MMc

 

$<var>y+33y = 918</var>$

 

$<var>y= \frac{918}{34}</var>$

 

$<var>y=27</var>$

 

$<var>3^x = 27</var>$

 

$<var>3^x = 3^3</var>$

 

$<var>\boxed{x=3}</var>$

 

Desculpe eu fiz errado na primeira vez.

Answer 2

$3^{x-1}+3^{x+1}+3^{x+2}=3^{x}+306$

$<var>3^{x-1}+3^{x+1}+3^{x+2}-306-3^x=0</var>$

$<var>\frac{34}{3}(3^x-27)=0</var>$

$<var>3^x-27=0</var>$

$<var>3^x=27</var>$

$<var>3^x=3^3</var>$

$<var>x=3</var>$