Solution.

(a)

\begin {align*} 6^{2x-1} & = 9^{x+3}\\ 6^{2x}6^{-1} & = 9^x\cdot 9^3\\ \implies \frac {6^{2x}}{6} & = (729)9^x \end {align*}

And it becomes very difficult to solve. So, \begin {align*} \log _66^{2x-1} & = \log _69^{x+3}\\\\ \implies (2x-1)\log _66 & = (x+3)\log _69\\\\ \implies 2x-1 & = (x+3)\log _69\\\\ \implies x(2-\log _69) & = 1+3\log _69\\\\ \implies x & = \frac {1+3\log _69}{2-\log _69}\\ \end {align*}

(b)

\(\log _3(2-3x)=\log _9(6x^2-19x+2)\)

Let us change the base of the R.H.S to base 3. \[\iff \log _9(6x^2-19x+2)=\frac {\log _3(6x^2-19x+2)}{\log _39}\] Thus, \(\log _3(2-3x)=\frac {\log _3(6x^2-19x+2)}{2\log _33}\)

\begin {align*} \implies 2\log _3(2-3x) & = \log _3(6x^2-19x+2)\\\\ \implies \log _3(2-3x)^2 & = \log _3(6x^2-19x+2)\\\\ \therefore (2-3x)^2 & = 6x^2-19x+2\\\\ \implies 4-12x+9x^2 & = 6x^2-19x+2\\\\ \therefore 3x^2+7x+2 & = 0\\\\ \implies x=\frac {-1}{3},-2\\ \end {align*}

Proof.

\[a^x=b^y\cdots (1)\]

\[b^y=(ab)^{xy}\cdots (2)\]

\[\text {Using (2)}\quad \implies \log b^y=(xy)\log (ab)\]

\[\log b^y=(xy)[\log a + \log b]\]

\[\log b^y = y \log a^x\]

\[\log b^y = xy \log a + xy \log b\]

\[\log b^y = y\log a^x+x\log b^y\]

\[\therefore \log b^y =y\log b^y+x\log b^y\]

\[\implies \log b^y = \log b^y(y+x)\]

\[\therefore 1=y+x\] □

Solution.

\[\log _3x\log _3y=-6\cdots (1)\]

\[\log _3x+\log _3y=\frac {5}{2}\cdots (2)\]

\[\implies \log _3x=\frac {5}{2}-\log _3y\]

\[\log _3x\log _3y=-6\]

\[\left (\frac {5}{2}-\log _3y\right )\log _3y=-6\]

\[\frac {5}{2}\log _3y-(\log _3y)^2=-6\]

\[\implies 2(\log _3y)^2-5\log _3y-12=0\]

\[\text {Let}\quad z=\log _3y.\quad \text {Then}\quad 2z^2-5z-12=0\]

\[\implies (2z+3)(z-4)=0\]

\[z=\frac {-3}{2}\quad \text {or}\quad 4\]

Thus,

\(\log _3y=4\implies y=3^4\implies y=81\)

\(\log _3y=\frac {-3}{2}\implies 3^{\frac {-3}{2}}=y\implies y=3^{\frac {1}{3/2}}\)

So, when \(y=81\)

\(\log _3x=\frac {5}{2}-\log _3y\)

\(\implies \log _3x=\frac {5}{2}-\log _381\)

\(\log _3x = \frac {5}{2}-4=\frac {-3}{2}\)

\(\log _3x=\frac {-3}{2}\implies x=3^{-3/2}\)

(a)

when \(y=3^{-3/2}\)

\(\log _3x=\frac {5}{2}-\log _33^{-3/2}\)

\(\implies \log _3x=\frac {5}{2}+\frac {3}{2}=4\)

\(\therefore \log _3x=4\implies x=3^4=81\)