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.

Related questions
See all questions in Logarithmic Models
Impact of this question
1421 views around the world
You can reuse this answer
Creative Commons License