How do you solve #8^(2x-5)=5^(x+1)#?

Question

↳Redirected from "Question #c42fa"

4949 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-10T14:03:40

#x = (5ln(8)+ln(5))/(2ln(8)-ln(5))~~4.7095#

Using the property of logarithms that #log(a^x) = xlog(a)#, we have

#8^(2x-5) = 5^(x+1)#

#=> ln(8^(2x-5)) = ln(5^(x+1))#

#=> (2x-5)ln(8) = (x+1)ln(5)#

#=> 2ln(8)x - 5ln(8) = ln(5)x+ln(5)#

#=>2ln(8)x - ln(5)x = 5ln(8)+ln(5)#

#=> (2ln(8)-ln(5))x = 5ln(8)+ln(5)#

#:. x = (5ln(8)+ln(5))/(2ln(8)-ln(5))~~4.7095#