Question

How do you solve ${\log _4}x - {\log _4}(x - 1) = \dfrac{1}{2}$?

Answer 1

Use special log properties, and solve for x to get the answer
Before beginning the above problem, we will have to use and remember 2 different log properties in this question. Which are$\log a - \log b = \log \dfrac{a}{b}$ and ${\log _b}a = c \Rightarrow {b^c} = a$. Using the first property in the first step we will reduce the whole LHS into 1 log term form. After this we will use the 2 nd log property which will finally give us an equation with a single variable. Solving this we will get our answer as $x = 2$.

Complete step by step solution:
The given question we have is ${\log _4}x - {\log _4}(x - 1) = \dfrac{1}{2}$
Now, we will use one log property which states that:-
$\log a - \log b = \log \dfrac{a}{b}$
Here,
$\log a = {\log _4}x$ and $\log b = {\log _4}(x - 1)$
Solving using the above property, we will get
${\log _4}\left( {\dfrac{x}{{x - 1}}} \right) = \dfrac{1}{2} \\\ \\\$ ${\log _4}\left( {\dfrac{x}{{x - 1}}} \right) = \dfrac{1}{2} \\\ \\\$
Using another property of log at this point to solve the given problem. We will get:-
${\log _b}a = c \Rightarrow {b^c} = a \\\ {\log _4}\dfrac{x}{{x - 1}} = \dfrac{1}{2} \\\ \Rightarrow {4^{\dfrac{1}{2}}} = \dfrac{x}{{x - 1}} \\\ \Rightarrow \sqrt 4 = \dfrac{x}{{x - 1}} \\\ \Rightarrow 2 = \dfrac{x}{{x - 1}} \\\$
Cross multiplying the above equation, we will get
$x = (x - 1) \times 2 \\\ \Rightarrow x = 2x - 2 \\\ \Rightarrow x - 2x = - 2 \\\ \Rightarrow - x = - 2 \\\ \Rightarrow x = 2 \\\$
Therefore, 2 is the solution of the given question

Note: Using which log property at what time so that you can get answers quickly is a pure skill. You will need to hone your solving skills by practicing more and more questions. Only after solving a lot of questions will you know what step or or what property to use so that we can get the best answer in the least possible steps.