If $3^{x}=3^{20} \cdot 3^{20} \cdot 3^{18}+3^{19} \cdot 3^{20} \cdot 3^{19}+3^{18} \cdot 3^{21} \cdot 3^{19}$, determine the value of $x$.
Solution 1

Using the rules for manipulating exponents,

$$
\begin{aligned}
3^{x} & =3^{20} \cdot 3^{20} \cdot 3^{18}+3^{19} \cdot 3^{20} \cdot 3^{19}+3^{18} \cdot 3^{21} \cdot 3^{19} \\
& =3^{20+20+18}+3^{19+20+19}+3^{18+21+19} \\
& =3^{58}+3^{58}+3^{58} \\
& =3\left(3^{58}\right) \\
& =3^{1+58} \\
& =3^{59}
\end{aligned}
$$

Since $3^{x}=3^{59}$, then $x=59$.

## Solution 2

We simplify the right side of the given equation by first removing a common factor which we obtain by taking the smallest of the three exponents of the first factors in each term, the smallest of the three exponents of the second factors, and the smallest of the three exponents of the third factors:

$$
\begin{aligned}
3^{x} & =3^{20} \cdot 3^{20} \cdot 3^{18}+3^{19} \cdot 3^{20} \cdot 3^{19}+3^{18} \cdot 3^{21} \cdot 3^{19} \\
& =3^{18} \cdot 3^{20} \cdot 3^{18}\left(3^{2} \cdot 3^{0} \cdot 3^{0}+3^{1} \cdot 3^{0} \cdot 3^{1}+3^{0} \cdot 3^{1} \cdot 3^{1}\right) \\
& =3^{18+20+18}(9 \cdot 1 \cdot 1+3 \cdot 1 \cdot 3+1 \cdot 3 \cdot 3) \\
& =3^{56}(9+9+9) \\
& =3^{56}(27) \\
& =3^{56} 3^{3} \\
& =3^{59}
\end{aligned}
$$

Since $3^{x}=3^{59}$, then $x=59$.

ANSWER: $x=59$