remember that [tex](a^b)^c=a^{bc}[/tex]
and [tex](ab)^c=(a^c)(b^c)[/tex]
and if [tex]a^b=c^b[/tex] if b=b≠0, then a=c
and [tex]ln(a^b)=b(ln(a))[/tex]
and ln(1)=0
[tex]100^{x+1}=25^{x+1}[/tex]
100=25*4
[tex](25*4)^{x+1}=25^{x+1}[/tex]
[tex](25^{x+1})(4^{x+1})=25^{x+1}[/tex]
I guess that's a common base?
if we were to solve for x then divide both sides by [tex]25^{x+1}[/tex]
[tex]4^{x+1}=1[/tex]
take the ln of both sides
[tex]ln(4^{x+1})=ln(1)[/tex]
[tex](x+1)ln(4)=0[/tex]
[tex](x+1)=0[/tex]
[tex]x=-1[/tex]
