Answer:
n = 7
Step-by-step explanation:
[tex]\sf \log_4(64)^{n+1}=log_5(625)^{n-1}[/tex]
Change 64 to an exponent with base 4 and 625 to an exponent with base 5:
[tex]\implies \sf \log_4(4^3)^{n+1}=log_5(5^4)^{n-1}[/tex]
Using exponent rule [tex](a^b)^c=a^{bc}[/tex]
[tex]\implies \sf \log_4(4)^{3(n+1)}=log_5(5)^{4(n-1)}[/tex]
Using log rule: [tex]\log_a(b^c)=c \log_a(b)[/tex]
[tex]\implies \sf 3(n+1)\log_4(4)}=4(n-1)log_5(5)[/tex]
Using log rule: [tex]\sf \log_a(a)=1[/tex]
[tex]\implies \sf 3(n+1)=4(n-1)[/tex]
[tex]\implies \sf 3n+3=4n-4[/tex]
[tex]\implies \sf 3+4=4n-3n[/tex]
[tex]\implies \sf n=7[/tex]
