Question

Solve the following equations, having given $\log 2,\log 3,$and $\log 7.$

{2^{x + y}} = {6^y} \\\ {3^x} = {3.2^{y + 1}} \\\ \end{gathered} \right\\}$$.

Answer 1

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.