1. If $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$, prove that $\log (a+c)+\log (a+c-2 b)=2 \log |a-c|$ . \begin{aligned} & \frac{2}{b}=\frac{1}{a}+\frac{1}{c} \\ & \therefore 2 a c=b(a+c) \quad \ldots \ldots .(1) \\ & \text { Now } \\ & \log (a+c)+\log (a+c-2 b) \\ & =\log (a+c)(a+c-2 b) \\ & =\log \left\{(a+c)^2-2 b(a+c)\right\} \\ & =\log \left\{(a+c)^2-4 a c\right\} \quad[\text { from (1)] } \\ & =\log (a-c)^2 \\ & =2 \log |a-c| \quad(\text { proved }) \end{aligned} 2. Prove that $a^{\log _{a^2}x} \times b^{\log _{b^2} y} \times c^{\log _{c^2}z}=\sqrt{ x y z}$. \begin{aligned} & \text{Solution:}\\ & a^{\log _{a^2}x} \times b^{\log _{b^2} y}\times c^{\log _{c^2}z}=P \\ & \Rightarrow \log _{a^2} x \log a+\log _{b^2}y \log b+\log _{c^2} z \log c=\log p \\ & \Rightarrow \frac{\log x}{\log a^2} \log a+\frac{\log y}{\log y^2} \log b+\frac{\log z}{\log c^2}\log c=\log p \\ & \Rightarrow \quad \frac{\log x}{2}+\frac{\log y}{2}+\frac{\log z}{2}=\log x \\ & \Rightarrow \quad \frac{1}{2} \log (x y z)=\log P\\ & \Rightarrow \sqrt{ x y z}=P \\ & \end{aligned} Second process $$ \begin{aligned} & a^{\log _{a^2} x}=p \\ \Rightarrow & \log _e a \log _{a^2} x=\log _e p \\ \Rightarrow & \log _e a \times \frac{\log _e x}{\log _e a^2}=\log _e p \\ \Rightarrow & \log _e a \times \frac{\log _e x}{2 \log _e a}=\log _e p \\ \Rightarrow & \log _e x=2 \log _e p \\ \Rightarrow & x=p^2 \\ \therefore & p=\sqrt{x} \end{aligned} $$ similarly $b^{\log _b y}=\sqrt{y}$ $\text {and}\ c^{\log _c z^z}=\sqrt{z}$ $$ \therefore a^{\log _{a^2}x} \times b^{\log _{b^2} y}\times c^{\log _{c^2}z}=\sqrt{x y z} $$ 3. Solve $x^{\log _{x^2}\left(x^2-1\right)}=5$ \begin{aligned} & x^{\log _{x^2}\left(x^2-1\right)}=5 \\ & \Rightarrow x^{\frac{\log _{x}\left({x^2-1}\right)}{\log _x x^2}}=5 \quad\left[\because \log _N M=\frac{\log _a M}{\log _a N}\right] \\ & \Rightarrow x^{\frac{\log _x\left(x^2-1\right)}{2 \log _x x}}=5 \quad\left[\because \log _a a=1\right] \\ & \Rightarrow x^{\frac{\log _x\left(x^2-1\right)}{2}}=5 \\ & \Rightarrow\left[x^{\log _x\left(x^2-1\right)}\right]^{\frac{1}{2}}=5 \quad\left[\because a^{\log _aM}=-M\right] \\ & \Rightarrow \quad\left(x^2-1\right)^{\frac{1}{2}}=5 \\ & \Rightarrow \quad x^2-1=25 \text {, (by squarieng bsth sides) } \\ & \Rightarrow x^2=26 \\ & \Rightarrow x=\sqrt{26} \\ & \end{aligned}