Question

How do you solve $\log 2x = \log 4$ ?

Answer 1

This problem of logarithm is solved by using the formula $\log _a^b - \log _a^c = \log _a^{\dfrac{b}{c}}$. We know that if $\log _a^b = y$ then $b = {a^y}$. First of all bring the term $\log 4$ on the left hand side of the given equation and then apply the above given formula and equating the obtained equation we get the required result.

Complete step by step answer:
Given, we have to find the particular value of $x$ so that it satisfied the given equation $\log 2x = \log 4$.
We have $\log 2x = \log 4$.
Taking all the terms on the left hand side of the equation we get,
$\Rightarrow \log 2x - \log 4 = 0$
Since the base of logarithm is not given so we usually suppose the base is $10$.
Now, applying the formula $\log _a^b - \log _a^c = \log _a^{\dfrac{b}{c}}$ we can write $\log _{10}^{2x} - \log _{10}^4 = 0$ as $\log _{10}^{\dfrac{{2x}}{4}}$.
$\Rightarrow \log _{10}^{2x} - \log _{10}^4 = 0 \\\ \Rightarrow \log _{10}^{\dfrac{{2x}}{4}} = 0$
we know that if $\log _a^b = y$ then we can write $b = {a^y}$. Applying this rule we can write
$\Rightarrow \dfrac{{2x}}{4} = {10^0}$
We have studied in the chapter exponent and power that if the power of any number other than zero is zero then its value is one. That is ${y^0} = 1$ where $y \ne 0$.
So, we get $\dfrac{{2x}}{4} = 1$
$\Rightarrow 2x = 4 \times 1 \\\ \Rightarrow x = \dfrac{4}{2} \\\ \therefore x = 2$

Thus, we get $x = 2$ which satisfies the given equation $\log 2x = \log 4$.

Note: This problem can simply be solved by equating $2x$ and $4$ because the bases of the logarithm are same but if more than two terms are present on either side of equation then firstly combine all the terms using logarithmic formula to make single terms of same base and then do the same.
Some other formula that are extremely important when working with the problem of logarithms are
(1) $\log _a^b + \log _a^c = \log _a^{b \times c}$.
(2) $\log _a^b \times \log _b^c \times \log _c^d = \log _a^d$.
(3) $\log _{{x^m}}^{{x^n}} = \dfrac{n}{m}$.