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

Answer 1

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

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.