How do you solve # 4^(x-1)=5^(x+1)#?

Question

4652 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-31T20:10:29

#4^x-1 = 4^-1*4^x = 4^x/4#

#5^(x+1) = 5^1*5^x = 5*5^x#

Given

#4^(x-1) = 5^(x+1)#

We have

#4^x/4 = 5*5^x#

Multiply both sides by #4# and divide both sides by #5^x# to get

#20 = 4^x/5^x = (4/5)^x#

Now take logs of both sides to get

#x*log(4/5) = log(20)#

Divide both sides by #log(4/5)# to get

#x = log(20)/log(4/5) = (log(4)+log(5))/(log(4)-log(5))#

#log(2) ~= 0.30103#

#log(4) = 2log(2) ~= 0.60206#

#log(5) = 1 - log(2) ~= 0.69897#

#x ~= (0.60206+0.69897)/(0.60206-0.69897)#

#=1.30103/-0.09691#

#~=-13.425#