\(3^{3 x - 1} = 3 \sqrt{3} = 3^{1} ⋅ 3^{\frac{1}{2}} = 3^{1\frac{1}{2}} \text{.}\)

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(3 x - 1 = 1\frac{1}{2} \text{.}\)

Balansmethode geeft \(x = \frac{5}{6} \text{.}\)

Balansmethode geeft \(3 ⋅ 4^{2 x - 3} = 12\) dus \(4^{2 x - 3} = 4 \text{.}\)

\(4 = 4^{1} \text{,}\) dus \(4^{2 x - 3} = 4^{1} \text{.}\)

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(2 x - 3 = 1 \text{.}\)

Balansmethode geeft \(x = 2 \text{.}\)

Grondtal gelijk maken geeft \(5^{1} ⋅ 5^{x} = (5^{-1})^{x + 3} \text{.}\)

Herleiden geeft \(5^{x + 1} = 5^{-x - 3} \text{.}\)

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(x + 1 = -x - 3 \text{.}\)

Balansmethode geeft \(x = -2 \text{.}\)

\(3^{x + 1} = 81 = 3^{4} \text{.}\)

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(x + 1 = 4\)
Balansmethode geeft \(x = 3 \text{.}\)

Uit de definitie van logaritme volgt \(-4 x - 4 = 2^{4} = 16 \text{.}\)

Balansmethode geeft \(-4 x = 20\) dus \(x = -5 \text{.}\)

Balansmethode geeft \({}^{3}\!\log(-2 x + 3) = 2 \text{.}\)

Uit de definitie van logaritme volgt \(-2 x + 3 = 3^{2} = 9 \text{.}\)

Balansmethode geeft \(-2 x = 6\) dus \(x = -3 \text{.}\)