How do you solve $81^{x} = 243^{x} + 2$ $81^x = 243^x + 2$ ?

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How do you solve $81^{x} = 243^{x} + 2$ $81^x = 243^x + 2$ ?

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Answer 1 · 2018-03-05T13:02:16

$There is no real solution for the equation.$ $"There is no real solution for the equation."$

Explanation:

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

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

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

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

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How do you solve $81^{x} = 243^{x} + 2$ $81^x = 243^x + 2$ ?

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