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

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

Balansmethode geeft \(x = 2\frac{1}{3} \text{.}\)

Balansmethode geeft \(3 ⋅ 2^{x - 3} = 48\) dus \(2^{x - 3} = 16 \text{.}\)

\(16 = 2^{4} \text{,}\) dus \(2^{x - 3} = 2^{4} \text{.}\)

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

Balansmethode geeft \(x = 7 \text{.}\)

Grondtal gelijk maken geeft \((5^{2})^{x + 3} = 5^{2} ⋅ 5^{x} \text{.}\)

Herleiden geeft \(5^{2 x + 6} = 5^{x + 2} \text{.}\)

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

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

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

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

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

Balansmethode geeft \(-2 x = 12\) dus \(x = -6 \text{.}\)

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

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

Balansmethode geeft \(-2 x = 18\) dus \(x = -9 \text{.}\)