Answer:
[tex]-4[/tex]
Step-by-step explanation:
The given logarithmic equation is:
[tex]\log_7(x+1)+\log_7x=\log_712[/tex]
Recall and apply product rule of logarithms.
[tex]\log_aM+\log_aN=\log_aMN[/tex]
We apply this property to the left side of the equation to get;
[tex]\log_7x(x+1)=\log_712[/tex]
We take the antilogarithm of both sides to get:
[tex]x(x+1)=12[/tex]
We expand to obtain:
[tex]x^2+x=12[/tex]
We rewrite in the standard quadratic form:
[tex]x^2+x-12=0[/tex]
We factor to obtain:
[tex](x-3)(x+4)=0[/tex]
Either [tex](x-3)=0[/tex] or [tex](x+4)=0[/tex]
Either [tex]x=3[/tex] or [tex]x=-4[/tex]
But the domain is[tex]x\:>\:0[/tex].
Hence [tex]x=-4[/tex] is an extraneous solution.
