\input a4
\def\frac#1#2{{#1\over#2}}
\nopagenumbers
\tabskip=1em plus 2em minus .5em
\newskip\ourskip
\ourskip=1.2cm
\halign to \hsize{\vrule width0pt height4ex depth3ex #\hfil&#\hfil&#\hfil\cr
$\frac14N = 8^{1\frac13x}$&$y = 3 + 4^{6x}$&$y = 3 + 8^{4x}$\cr
\noalign{\vskip\ourskip}
$N = 4\cdot 8^{1\frac13x}$&$y = 3 + 8^{{}^8\!\log(4)\cdot 6x}$&$y - 3 = 4^{6x}$\cr
\noalign{\vskip\ourskip}
$N = 8^{{}^8\!\log(4)}\cdot 8^{1\frac13x}$&$y = 3 + 4^{{}^4\!\log(8)\cdot 4x}$&$x = \frac16\cdot{}^4\!\log(y-3)$\cr
\noalign{\vskip\ourskip}
$N = 8^{\frac23}\cdot 8^{1\frac13x}$&$y - 3 =  8^{4x}$&$x=\frac14\cdot{}^8\!\log(y-3)$\cr
\noalign{\vskip\ourskip}
$N = 8^{1\frac13x+\frac23}$&$4x = {}^8\!\log(y-3)$&${}^g\!\log(a)+{}^g\!\log(b) = A$\cr
\noalign{\vskip\ourskip}
$1\frac13x+\frac23 = {}^8\!\log(N)$&$6x = \frac1{{}^8\!\log(4)}\cdot{}^8\!\log(y-3)$&$g^{{}^g\!\log(a)+{}^g\!\log(b)} = g^A$\cr
\noalign{\vskip\ourskip}
$1\frac13x = {}^8\!\log(N)-\frac23$&$4x = \frac1{{}^4\log(8)}\cdot{}^4\!\log(y-3)$&$g^{{}^g\!\log(a)}\cdot g^{{}^g\!\log(b)} = g^A$\cr
\noalign{\vskip\ourskip}
$1\frac13x = {}^8\!\log(N)-{}^8\!\log(8^\frac23)$&${}^8\!\log(4)\cdot 6x = {}^8\!\log(y-3)$&$a\cdot b = g^A$\cr
\noalign{\vskip\ourskip}
$1\frac13x = {}^8\!\log(N)-{}^8\!\log(4)$&${}^4\!\log(8)\cdot 4x = {}^4\!\log(y-3)$&${}^g\!\log(a\cdot b) = A$\cr
\noalign{\vskip\ourskip}
$1\frac13x = {}^8\!\log(\frac14N)$&$6x = {}^4\!\log(y-3)$\cr
}
\bye
