remember
[tex]log_a(b)=c[/tex] translates to [tex]a^c=b[/tex]
so
[tex]log_x(\frac{1}{32})=-5[/tex] translates to [tex]x^{-5}=\frac{1}{32}[/tex]
remember, if a^b=c^b when b=b, then a=c
and x^-m=1/(x^m)
hmm
[tex]x^{-5}=\frac{1}{32}[/tex]
[tex]\frac{1}{x^5}=\frac{1}{32}[/tex]
32=x⁵
5th root both sides
2=x
x=2
