Answer

Verified

471k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Which are the Top 10 Largest Countries of the World?

Write a letter to the principal requesting him to grant class 10 english CBSE

10 examples of evaporation in daily life with explanations

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE