Question

HELP I need to find at least the common base: 100^x +1 = 25^x +1

Answer 1

remember that
`(a^b)^c=a^(bc)`

and
`(ab)^c=(a^c)(b^c)`

and if
`a^b=c^b` if b=b≠0, then a=c

and
`ln(a^b)=b(ln(a))`

and ln(1)=0

`100^(x+1)=25^(x+1)`

100=25*4

`(25*4)^(x+1)=25^(x+1)`

`(25^(x+1))(4^(x+1))=25^(x+1)`

I guess that's a common base?

if we were to solve for x then divide both sides by
`25^(x+1)`

`4^(x+1)=1`

take the ln of both sides

`ln(4^(x+1))=ln(1)`

`(x+1)ln(4)=0`

`(x+1)=0`

`x=-1`