# How do you solve 81^x = 243^x + 2?

Mar 5, 2018

$\text{There is no real solution for the equation.}$

#### Explanation:

$243 = 3 \cdot 81$
$\implies {81}^{x} = {\left(3 \cdot 81\right)}^{x} + 2$
$\implies {81}^{x} = {3}^{x} \cdot {81}^{x} + 2$
$\implies {81}^{x} \left(1 - {3}^{x}\right) = 2$
$\implies {\left({3}^{x}\right)}^{4} \left(1 - {3}^{x}\right) = 2$
$\text{Name "y = 3^x", then we have}$
$\implies {y}^{4} \left(1 - y\right) = 2$
$\implies {y}^{5} - {y}^{4} + 2 = 0$
$\text{This quintic equation has the simple rational root "y= -1.}$
$\text{So "(y+1)" is a factor, we divide it away :}$
$\implies \left(y + 1\right) \left({y}^{4} - 2 {y}^{3} + 2 {y}^{2} - 2 y + 2\right) = 0$
$\text{It turns out that the remaining quartic equation has no real}$ $\text{roots. So we have no solution as "y = 3^x > 0" so } y = - 1$
$\text{does not yield a solution for } x .$

$\text{Another way to see that there is no real solution is :}$
${243}^{x} \ge {81}^{x} \text{ for positive "x", so "x" must be negative.}$
$\text{Now put "x = -y" with "y" positive, then we have}$

${\left(\frac{1}{243}\right)}^{y} + 2 = {\left(\frac{1}{81}\right)}^{y}$

$\text{but "0 <= (1/243)^y <= 1" and } 0 \le {\left(\frac{1}{81}\right)}^{y} \le 1$
$\text{So " (1/243)^y + 2 " is always bigger than } {\left(\frac{1}{81}\right)}^{y} .$