How do you solve $243^(.2x)=81^(x+5)$ ?

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Answer 1 · 2016-03-11T13:45:48

$x = -20 /3$

$243 ^(.2x) = 81^ (x+5)$

Simplifying the bases by prime factorisation:

$243 =3*3*3*3*3 =color(blue)( 3^5$

$81 = 3*3*3*3 = color(blue)(3^4$

Thus, using the rule $(a^b)^c=a^((bc))$ :

$243 ^(,2x) = 3^((5xx.2x)) = 3^x$

$81^ (x+5) = 3^((4(x+5))) = 3^(4x + 20)$

The expression becomes:

$3^ (x) = 3^(4x + 20)$

The bases are equal so we can equate powers:

$x= 4x + 20$

$x-4x= 20$

$-3x = 20$

$x = -20/3$

How do you solve 243^(.2x)=81^(x+5)243^(.2x)=81^(x+5)?