# How do you solve #2^x=7^(x-1)#?

##### 1 Answer

May 30, 2016

#### Explanation:

Dividing both sides by

#1/7 = 2^x/7^x = (2/7)^x#

Then taking logs (any base):

#log(1/7) = log((2/7)^x) = x log(2/7)#

Divide both sides by

#x = log(1/7)/log(2/7) = (-log(7))/(log(2)-log(7)) = log(7)/(log(7)-log(2)) ~~ 1.5533#

**Complex solutions**

We can find all Complex solutions by introducing a multiplier

#1/7 = (2/7)^x * e^(2kpii)#

#ln(1/7) = ln((2/7)^x * e^(2kpii)) = x ln(2/7) + 2kpii#

#x ln(2/7) = ln(1/7)-2kpii#

#x = (ln(1/7)-2kpii)/ln(2/7) = (-ln(7)-2kpii)/(ln(2)-ln(7)) = (ln(7)+2kpii)/(ln(7)-ln(2))#

where