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#