\(5^{2 x + 1} = {1 \over 5} \sqrt{5} = 5^{-1} ⋅ 5^{\frac{1}{2}} = 5^{-\frac{1}{2}} \text{.}\)

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

Balansmethode geeft \(x = -\frac{3}{4} \text{.}\)

Balansmethode geeft \(4 ⋅ 2^{3 x + 2} = 64\) dus \(2^{3 x + 2} = 16 \text{.}\)

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

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(3 x + 2 = 4 \text{.}\)

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

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

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

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

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

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

\(g^{A} = g^{B}\) geeft \(A = B \text{,}\) dus \(x + 4 = 4\)
Balansmethode geeft \(x = 0 \text{.}\)

Uit de definitie van logaritme volgt \(3 x - 1 = 2^{3} = 8 \text{.}\)

Balansmethode geeft \(3 x = 9\) dus \(x = 3 \text{.}\)

Balansmethode geeft \({}^{2}\!\log(-3 x - 5) = 0 \text{.}\)

Uit de definitie van logaritme volgt \(-3 x - 5 = 2^{0} = 1 \text{.}\)

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