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

$ln(4^{x+1})=ln(1)$

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

$(x+1)=0$

$x=-1$