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Solve the following equations, having given \[\log 2,\log 3,\]and \[\log 7.\]
\[\left\{ \begin{gathered}
  {2^{x + y}} = {6^y} \\
  {3^x} = {3.2^{y + 1}} \\
\end{gathered} \right\}\].

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Hint- Use Product rule of logarithm \[{\log _a}\left( {m \times n} \right) = {\log _a}m + {\log _a}n\],Power rule of logarithm\[{\text{lo}}{{\text{g}}_a}{m^n} = n{\log _a}m\]and Division rule of Logarithm\[{\log _a}\dfrac{m}{n} = {\log _a}m - {\log _a}n\].

In this question we need to use basic rules of logarithm and solve the given equation. Now, as per question, we have
\[{{\text{2}}^{x + y}} = {6^y};{3^x} = {3.2^{y + 1}}\]
Now, to obtain$x$value we proceed as
\[{{\text{2}}^{x + y}} = {\left( {2.3} \right)^y} \Rightarrow {2^x}{.2^y} = {2^y}{.3^y}\]
Cancelling \[{2^y}\]from both sides, we get
\[ \Rightarrow {2^x} = {3^y}\]
Applying $\log $ on both sides and using power rule of logarithm bring the power downwards, we get
\[
  x\log 2 = y\log 3 \\
   \Rightarrow y = \dfrac{{x\log 2}}{{\log 3}} \\
\]
Similarly for \[{3^x} = {3.2^{y + 1}}\]
If we rearrange the above equation and taking $3$ on RHS, we get
\[{3^{x - 1}} = {2^{y + 1}}\]
Applying $\log $ on both sides and use Power rule of logarithm, we get
\[\left( {x - 1} \right)\log 3 = \left( {y + 1} \right)\log 2\]
By substituting, \[y = \dfrac{{x\log 2}}{{\log 3}}\] obtained previously in above expression
\[\left( {x - 1} \right)\log 3 = \left( {\dfrac{{x\log 2}}{{\log 3}} + 1} \right)\log 2\]
Taking LCM and Cross multiplying $\log 3$, we get
\[ \Rightarrow x{\left( {\log 3} \right)^2} - {\left( {\log 3} \right)^2} = x{\left( {\log 2} \right)^2} + \log 2\log 3\]
Now rearrange the above expression to find$x$value
\[ \Rightarrow x = \dfrac{{\log 3\left( {\log 3 + \log 2} \right)}}{{{{\left( {\log 3} \right)}^2} - {{\left( {\log 2} \right)}^2}}};\] and we know \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
So \[x = \dfrac{{\log 3\left( {\log 3 + \log 2} \right)}}{{\left( {\log 3 + \log 2} \right)\left( {\log 3 - \log 2} \right)}};\]
Cancelling \[\left( {\log 3 + \log 2} \right)\]from numerator and denominator, we get
\[x = \dfrac{{\log 3}}{{\log 3 - \log 2}}\] and So the value of\[y = \dfrac{{\log 2}}{{\log 3 - \log 2}}\]

Note- Whenever this type of question appears always first note down the given things. Afterwards resolve and simplify the expression as much as possible. Using the power rule of logarithm \[{\text{lo}}{{\text{g}}_a}{m^n} = n{\log _a}m\] bring power downwards. Grasp the knowledge of Logarithm identities as it helps to solve the questions easily. While applying power rule do not confuse \[{\left( {\log 3} \right)^2} \ne {\text{log }}{{\text{3}}^2}\]both are different expressions.
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