\(3^{2 x - 1} = {1 \over 9} \sqrt[3]{3} = 3^{-2} ⋅ 3^{\frac{1}{3}} = 3^{-1\frac{2}{3}} \text{.}\)

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

Balansmethode geeft \(x = -\frac{1}{3} \text{.}\)

Balansmethode geeft \(3 ⋅ 5^{x - 2} = 1\,875\) dus \(5^{x - 2} = 625 \text{.}\)

\(625 = 5^{4} \text{,}\) dus \(5^{x - 2} = 5^{4} \text{.}\)

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(x - 2 = 4 \text{.}\)

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

Grondtal gelijk maken geeft \(4^{1} ⋅ 4^{x} = (4^{2})^{x + 3} \text{.}\)

Herleiden geeft \(4^{x + 1} = 4^{2 x + 6} \text{.}\)

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(x + 1 = 2 x + 6 \text{.}\)

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

\(3^{x + 1} = 9 = 3^{2} \text{.}\)

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

Uit de definitie van logaritme volgt \(-2 x - 3 = 4^{0} = 1 \text{.}\)

Balansmethode geeft \(-2 x = 4\) dus \(x = -2 \text{.}\)

Balansmethode geeft \({}^{4}\!\log(-5 x - 4) = 2 \text{.}\)

Uit de definitie van logaritme volgt \(-5 x - 4 = 4^{2} = 16 \text{.}\)

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