Solve for x: 5^{log x} + 3^{log x} = 3^{log x + 1} - 5^{log x - 1}.

Given,

5^{log x} + 3^{log x} = 3^{log x + 1} - 5^{log x - 1}

⇒ 5^{log x} + 3^{log x} = 3^{log x}.3¹ - 5^{log x}.5^-1

⇒ 5^{log x} + 5^{log x}.5^-1 = 3^{log x}.3¹ - 3^{log x}

⇒ 5^{log x}(1 + 5^-1) = 3^{log x}(3 - 1)

⇒ 5^{log x}$\Big(1 + \dfrac{1}{5}\Big)$ = 2.3^{log x}

⇒ 5^{log x}.$\dfrac{6}{5}$ = 2.3^{log x}

$\Rightarrow \dfrac{5^{\text{log } x}}{3^{\text{log } x}} = \dfrac{5 \times 2}{6} \\[1em] \Rightarrow \Big(\dfrac{5}{3}\Big)^{\text{log } x} = \dfrac{5}{3} \\[1em] \Rightarrow \text{log } x = 1 \\[1em] \Rightarrow \text{log } x = \text{log } 10 \\[1em] \Rightarrow x = 10.$

Hence, x = 10.