Question

How do you solve `5^(x+2) = 4^(1-x)`?

Answer 1

One easy way is using logarithms because there is one law of logarithms which states that `loga^n=n*loga`

So, let's put logarithms on both sides (to keep the equality):

`log5^(x+2)=log4^(1-x)`
`(x+2)log5=(1-x)log4`

Approximating `log5~=0.7` and `log4~=0.6`, we get

`(x+2)0.7=(1-x)0.6`
`0.7x+1.4=0.6-0.6x`
`1.3x=-0.8`
`x=-0.8/1.3~=-0.61`

Answer 2

Another way to do that would be
`5^(x+2)=4^(1-x)`
`5^x cdot 5^2=4^1 cdot 4^(-1)`
`25 cdot 5^x = 4 cdot 1/4^x`
`5^x cdot 4^x=4/25`
`20^x=0.16`
using the natural log
`log 20^x=log0.16`
`x log 20=log 0.16`
`x=(log 0.16)/(log 20) approx -0.612`
or using the log base 20
`log_(20) 20^x=log_(20) 0.16`
`x=log_(20) 0.16 approx -0.612`