plz ans for both

plz ans for both . logc -+log(z 21) —— = log — log c + log2 x (z+-)-z log (x+{ —3=0.

Dear student,
The solution to your 1st query has been provided below:

$$\begin{gathered} Given, \\ \log \left(x\right)-\frac{1}{2}\log \left(x-\frac{1}{2}\right)=\log \left(x+\frac{1}{2}\right)-\frac{1}{2}\log \left(x+\frac{1}{8}\right) \to \overline{)1} \\ \Rightarrow \log \left(x\right)-\log {\left(x-\frac{1}{2}\right)}^{\frac{1}{2}}=\log \left(x+\frac{1}{2}\right)-\log {\left(x+\frac{1}{8}\right)}^{\frac{1}{2}} \left(a \log \left(b\right)=\log (b{)}^a\right) \\ \Rightarrow \log \left(x\right)-\log \left(x+\frac{1}{2}\right)=\log {\left(x-\frac{1}{2}\right)}^{\frac{1}{2}}-\log {\left(x+\frac{1}{8}\right)}^{\frac{1}{2}} \\ \Rightarrow \log \left(\frac{x}{x+\frac{1}{2}}\right)=\log {\left(\frac{x-\frac{1}{2}}{x+\frac{1}{8}}\right)}^{\frac{1}{2}} \left[ \log \left(a\right)-\log \left(b\right)=\log \left(\frac{a}{b}\right)\right] \\ Cancelling \log on both sides, \\ \Rightarrow \frac{x}{x+\frac{1}{2}}={\left(\frac{x-\frac{1}{2}}{x+\frac{1}{8}}\right)}^{\frac{1}{2}} \\ Squaring on both sides, \\ \Rightarrow {\left(\frac{x}{x+\frac{1}{2}}\right)}^2=\frac{x-\frac{1}{2}}{x+\frac{1}{8}} \\ \Rightarrow \frac{{\left(x\right)}^2}{{\left(x+\frac{1}{2}\right)}^2}=\frac{x-\frac{1}{2}}{x+\frac{1}{8}} \\ \Rightarrow x^2\left(x+\frac{1}{8}\right) = \left(x-\frac{1}{2}\right){\left(x+\frac{1}{2}\right)}^2 \\ \Rightarrow x^2\left(x+\frac{1}{8}\right) = \left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right) \\ \Rightarrow x^3+\left(\frac{1}{8}\right)x= \left(x^2 -\frac{1}{4}\right)\left(x+\frac{1}{2}\right) \\ \Rightarrow x^3+\left(\frac{1}{8}\right)x= x^3+\left(\frac{1}{2}\right)x^2 -\left(\frac{1}{4}\right)x-\frac{1}{8} \\ \Rightarrow \left(\frac{1}{8}\right)x=\left(\frac{1}{2}\right)x^2 -\left(\frac{1}{4}\right)x-\frac{1}{8} \\ \Rightarrow \left(\frac{1}{8}\right)x+\left(\frac{1}{4}\right)x=\left(\frac{1}{2}\right)x^2 -\frac{1}{8} \\ \Rightarrow \left(\frac{3}{8}\right)x=\left(\frac{4}{8}\right)x^2 -\frac{1}{8} \\ \Rightarrow 4x^2 -\; 3x-\; 1\; =\; 0 \\ \Rightarrow 4x^2 -4x+x-1\; =\; 0 \\ \Rightarrow 4x(x-1)\; +1(x-1)\; =\; 0 \\ \Rightarrow (4x+1\left)\right(x-1)=0 \\ \Rightarrow x=\; -\frac{1}{4} or x=1 \\ F\mathrm{or} \log \mathrm{to} \mathrm{be} \mathrm{defined} \\ x>0 \\ x>\frac{1}{2} \\ x>\; -\frac{1}{2} \\ x>\; -\frac{1}{8} (All terms inside \log in \overline{)1} must be greater than 0) \\ C\mathrm{ommon} \mathrm{value} \mathrm{of} \mathrm{this} is x>\frac{1}{2} \\ So, x\ne -\frac{1}{2} \\ H\mathrm{ence}, x=\; 1 \end{gathered}$$
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