How do you solve for x in #18^(x-2) = 13^(-2x)#?

1 Answer
Lovecraft
Sep 28, 2015

#x = (2ln(18))/(ln(18) +2ln(13))#

Explanation:

Since 13 is a prime and 18 is a compound number without 13 as a prime factor you'll never be able to equate bases, therefore, we'll need to apply logs. I'm using the natural base due to preference you can use whatever base you want - that being said, I'd recommend using natural or decimal.

#ln(18^(x-2)) = ln(13^(-2x))#
#(x-2)ln(18) = (-2x)ln(13)#
#xln(18) -2ln(18) = -2xln(13)#

Isolating x

#xln(18) +2xln(13) = 2ln(18)#
#x*(ln(18) + 2ln(13)) = 2ln(18)#
#x = (2ln(18))/(ln(18) +2ln(13))#

You could break down some of these logs further but since the problem gave 18 and 13 as numbers, I'd leave the answer in terms of their logs.

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