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

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Answer 1 · 2015-08-20T13:23:14

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

$3^(x^2+20)=(1/27)^(3x)$

We know that $1/27=1/(3^3$

So,

$3^(x^2+20)=(1/3^3)^(3x)$

By property
$color(blue)(1/a=a^-1$

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

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

$x^2+20=-9x$

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

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

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

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

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

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

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

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

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