Question

How do you solve $\log \left( {5x} \right) + \log \left( {x - 1} \right) = 2$?

Answer 1

Hint: In this question, we have to find the $x$ value from the above logarithmic equation. For that, we are going to simplify the equation. Next, we use some laws of logarithms and then simplify to arrive at our final answer. Next, check whether the values obtained are satisfying expressions or not. Finally, we get the required answer.

Formula used: Product rule: The logarithm of a product is the sum of the logarithm of the factors.
$\log \left( {a \times b} \right) = \log a + \log b$

Complete step-by-step solution:
We have been given that,
$\log \left( {5x} \right) + \log \left( {x - 1} \right) = 2$...........….. (1)
We know that, if $a > 0,b > 0$ then we have,
$\log a + \log b = \log ab$
Using the above rule in equation (1), we get
$ \Rightarrow \log \left( {5x} \right)\left( {x - 1} \right) = 2$
Open the brackets and multiply the terms,
$ \Rightarrow \log \left( {5{x^2} - 5x} \right) = 2$
We know that, if ${\log _b}a = c$, then $a = {b^c}$
Here, $a = 5{x^2} - 5x,b = 10$ and $c = 2$. Substituting the values in the expression, we get
$ \Rightarrow 5{x^2} - 5x = {10^2}$
Square the term on the right side,
$ \Rightarrow 5{x^2} - 5x = 100$
Take 5 commons from the left side,
$ \Rightarrow 5\left( {{x^2} - x} \right) = 100$
Divide both sides by 5,
$ \Rightarrow {x^2} - x = 20$
Move constant part on the left side,
$ \Rightarrow {x^2} - x - 20 = 0$
Now, find two factors of -20 in such a way that when those factors are added or subtracted, we get -1. The two such factors can be -5 and 4.
$ \Rightarrow {x^2} - 5x + 4x - 20 = 0$
Now, take common from the terms,
$ \Rightarrow x\left( {x - 5} \right) + 4\left( {x - 5} \right) = 0$
Again, take commonly from the terms,
$ \Rightarrow \left( {x - 5} \right)\left( {x + 4} \right) = 0$
Now equate each term with 0,
$ \Rightarrow x - 5 = 0$ and $x + 4 = 0$
Move the constant part on the right side,
$ \Rightarrow x = 5$ and $x = - 4$
Since the value of the log cannot be negative. So,
$ \Rightarrow x = 5$

Hence, the value of $x$ is 5.

Note: Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$