How do you solve $3^{x^{2} + 20} = {(\frac{1}{27})}^{3 x}$ $3^(x^2 + 20) = (1/27)^(3x)$ ?

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Answer 1 · 2015-08-20T13:23:14

The solutions are
$x = - 4, x = - 5$ $color(blue)(x=-4,x=-5$

$3^{x^{2} + 20} = {(\frac{1}{27})}^{3 x}$ $3^(x^2+20)=(1/27)^(3x)$

We know that $\frac{1}{27} = \frac{1}{3^{3}}$ $1/27=1/(3^3$

So,

$3^{x^{2} + 20} = {(\frac{1}{3^{3}})}^{3 x}$ $3^(x^2+20)=(1/3^3)^(3x)$

By property
$\frac{1}{a} = a^{- 1}$ $color(blue)(1/a=a^-1$

$3^{x^{2} + 20} = {(3^{- 3})}^{3 x}$ $3^(x^2+20)=color(blue)((3^-3))^(3x)$

$3^{x^{2} + 20} = 3^{- 9 x}$ $3^(x^2+20)=3^(-9x)$
Now as bases are equal we equate powers and find $x$ $x$

$x^{2} + 20 = - 9 x$ $x^2+20=-9x$

$x^{2} + 9 x + 20 = 0$ $x^2+9x+20=0$

We can Split the Middle Term of this expression to factorise it and find solutions.

$x^{2} + 9 x + 20 = 0$ $x^2+color(blue)(9x)+20=0$

$x^{2} + 5 x + 4 x + 20 = 0$ $x^2+color(blue)(5x+4x)+20=0$

$x (x + 5) + 4 (x + 5) = 0$ $x(x+5)+4(x+5)=0$

$(x + 4) (x + 5) = 0$ $(x+4)(x+5)=0$

We now equate the factors to zero.
$x + 4 = 0, x = - 4$ $x+4=0, x=-4$

$x + 5 = 0, x = - 5$ $x+5=0, x=-5$

How do you solve 3^(x^2 + 20) = (1/27)^(3x)3x2+20=(127)3x3^(x^2 + 20) = (1/27)^(3x)?