Question
solving exponential equations with exponents on each side
5^x =4^(x+1)
They don't have common bases and now I am lost I thought I could just come up with a common base (in this case 20) and change the equation to
20^4x =20^(5x+5)
However I am not getting the correct answer!
5^x =4^(x+1)
They don't have common bases and now I am lost I thought I could just come up with a common base (in this case 20) and change the equation to
20^4x =20^(5x+5)
However I am not getting the correct answer!
Answers
Reiny
Major Error !
How did you mathemagically change 5^x to 20^4x ?
Did you do something like (5^x)(4^4) = 20^4x ??
If so then something like this should work :
(3^5)(4^2) = 12^10 which is certainly false
You can only do this type of question if you know logs.
take the log of both sides
log(5^x) = log(4^(x+1))
using the rules of logs,
x(log5) = (x+1)(log4)
x(log5) = xlog4 + log4
xlog5 - xlog4 = log4
x(log5 - log4) = log4
x = log4/(log5 - log4)
= 6.21257
How did you mathemagically change 5^x to 20^4x ?
Did you do something like (5^x)(4^4) = 20^4x ??
If so then something like this should work :
(3^5)(4^2) = 12^10 which is certainly false
You can only do this type of question if you know logs.
take the log of both sides
log(5^x) = log(4^(x+1))
using the rules of logs,
x(log5) = (x+1)(log4)
x(log5) = xlog4 + log4
xlog5 - xlog4 = log4
x(log5 - log4) = log4
x = log4/(log5 - log4)
= 6.21257
Shellie
Thanks! I'm sorry for sooo many questions but I took this as a TERM class and so I've had 4 weeks to learn basically 1/2 the book with the final on Monday and he never really got time to lecture on Chapter 4 so I am basically trying to learn it on my own with your help and the help of videos on line!