Question

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

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$${\log }_a{\displaystyle \frac{m}{n}}={\log }_{a}m-{\log }_{a}n$$.

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
$$\begin{gathered} x\log 2=y\log 3 \\ \Rightarrow y={\displaystyle \frac{x\log 2}{\log 3}} \end{gathered}$$
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, $$y={\displaystyle \frac{x\log 2}{\log 3}}$$ obtained previously in above expression
$$\left(x-1\right)\log 3=\left({\displaystyle \frac{x\log 2}{\log 3}}+1\right)\log 2$$
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
$$\Rightarrow x={\displaystyle \frac{\log 3\left(\log 3+\log 2\right)}{{\left(\log 3\right)}^2-{\left(\log 2\right)}^2}};$$ and we know $$a^2-b^2=\left(a+b\right)\left(a-b\right)$$
So $$x={\displaystyle \frac{\log 3\left(\log 3+\log 2\right)}{\left(\log 3+\log 2\right)\left(\log 3-\log 2\right)}};$$
Cancelling $$\left(\log 3+\log 2\right)$$from numerator and denominator, we get
$$x={\displaystyle \frac{\log 3}{\log 3-\log 2}}$$ and So the value of$$y={\displaystyle \frac{\log 2}{\log 3-\log 2}}$$

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.