# 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\}\].

Answer

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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.

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.

Last updated date: 30th May 2023

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