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**Soroban** Hello, NameIsHidden!

We have: .$\displaystyle b \;=\;\frac{a^x + a^{-x}}{2}$

Multiply by 2: .$\displaystyle 2b \;=\;a^x + a^{-x}$

Multiply by $\displaystyle a^x\!:\;2ba^x \;=\;a^{2x} + 1 \quad\Rightarrowquad a^{2x} - 2ba^x + 1 \;=\;0$

Quadratic Formula: .$\displaystyle a^x \;=\;\frac{2b \pm\sqrt{4b^2-4}}{2} \quad\Rightarrow\quad a^x \;=\;b \pm\sqrt{b^2-1}$

Take logs, base $\displaystyle a$: .$\displaystyle \log_a(a^x) \;=\;\log_a(b \pm \sqrt{b^2-1)$

Therefore: .$\displaystyle x \;=\;\log_a(b+ \sqrt{b-1})\qquad \text{answer (b)}$