Question

Solve the algebraic expression for $$x:{\left(2x\right)}^{{\log }_{b}2}={\left(3x\right)}^{{\log }_{b}3}$$.
(a) $$x={\displaystyle \frac{1}{3}}$$
(b) $$x={\displaystyle \frac{1}{16}}$$
(c) $$x={\displaystyle \frac{1}{6}}$$
(d) $$x={\displaystyle \frac{1}{4}}$$

Answer 1

Hint: Take log to the base b, i.e., $${\log }_b$$, both the sides and use the formula $$\log \left(m^n\right)=n\log m$$ to simplify the expression. Now, use the product to sum-rule the logarithm given as: - $$\log \left(mn\right)=\log m+\log n$$ for further simplification. Use the algebraic identity: - $$(a^2-b^2)=(a+b)(a-b)$$ and cancel the like terms to get the answer.

Complete step by step solution:
Here, we have been provided with the logarithmic equation: - $${\left(2x\right)}^{{\log }_{b}2}={\left(3x\right)}^{{\log }_{b}3}$$ and we are asked to find the value of x.
$$\because {\left(2x\right)}^{{\log }_{b}2}={\left(3x\right)}^{{\log }_{b}3}$$
Taking log to the base b, i.e., $${\log }_b$$ both the sides and using the formula: - $$\log \left(m^n\right)=n\log m$$, we get,
$$\begin{array}{rl}& \Rightarrow {\log }_b{\left(2x\right)}^{{\log }_{b}2}={\log }_b{\left(3x\right)}^{{\log }_{b}3}\\ & \Rightarrow {\log }_{b}2.{\log }_b\left(2x\right)={\log }_{b}3.{\log }_b\left(3x\right)\end{array}$$
Using the product to sum of logarithm given by the formula $$\log \left(mn\right)=\log m+\log n$$, we get,
$$\begin{array}{rl}& \Rightarrow {\log }_{b}2.[{\log }_{b}2+{\log }_{b}x]={\log }_{b}3.[{\log }_{b}3+{\log }_{b}x]\\ & \Rightarrow {\left({\log }_{b}2\right)}^2+{\log }_{b}2.{\log }_{b}x={\left({\log }_{b}3\right)}^2+{\log }_{b}3.{\log }_{b}x\end{array}$$
Rearranging the terms we can write the above expression as: -
$$\Rightarrow {\log }_{b}2.{\log }_{b}x-{\log }_{b}3.{\log }_{b}x={\left({\log }_{b}3\right)}^2-{\left({\log }_{b}2\right)}^2$$
Using the algebraic identity: - $$(a^2-b^2)=(a+b)(a-b)$$, we get,
$$\Rightarrow {\log }_{b}x({\log }_{b}2-{\log }_{b}3)=({\log }_{b}3-{\log }_{b}2)({\log }_{b}3+{\log }_{b}2)$$
Taking (-1) common in the L.H.S., we have,
$$\Rightarrow -{\log }_{b}x({\log }_{b}3-{\log }_{b}2)=({\log }_{b}3-{\log }_{b}2)({\log }_{b}3+{\log }_{b}2)$$
Cancelling the common factors on both the sides, we get,
$$\begin{array}{rl}& \Rightarrow -{\log }_{b}x=({\log }_{b}3+{\log }_{b}2)\\ & \Rightarrow -{\log }_{b}x={\log }_b(3\times 2)\\ & \Rightarrow -{\log }_{b}x={\log }_{b}6\\ & \Rightarrow -{\log }_b\left(x^{-1}\right)={\log }_{b}6\end{array}$$
Using the formula $$a^{-m}={\displaystyle \frac{1}{a^m}}$$, we can write the above expression as: -
$$\Rightarrow {\log }_b\left({\displaystyle \frac{1}{x}}\right)={\log }_{b}6$$
Now, we can see that the base of the logarithm on both sides are the same, so we can equate their arguments by removing the log function. Therefore, we have,
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{x}}=6\\ & \Rightarrow x={\displaystyle \frac{1}{6}}\end{array}$$
Hence, option (c) is our answer.

Note: One may note that it will be difficult for us to solve the question in the exponential form and that is why we took the help of log to solve the question. The reason for taking log to the base ‘b’ was that we can clearly see that we had $${\log }_b$$ as the exponent on both sides of the original equation. Even if you will take any other base value you will need the base change rule given as: - $${\log }_{x}y={\displaystyle \frac{{\log }_{z}y}{{\log }_{z}x}}$$ to convert it into base ‘b’ at last.