Question

How do you solve $\log x + \log \left( {x - 9} \right) = 1$?

Answer 1

In this question we have solve the equation for $x$, which is a logarithmic function, so, the given question can solved by using properties of logarithms i.e., $\ln x + \ln y = \ln \left( {xy} \right)$, and $\log 10 = 1$, then solve by factoring the equation to get the required result.

Complete step by step solution:
Given equation is $\log x + \log \left( {x - 9} \right) = 1$,
Now using the fact that $\log 10 = 1$, and rewriting 1 as log 10, we get,
$\Rightarrow \log x + \log \left( {x - 9} \right) = \log 10$,
Now applying the logarithmic identity$\ln x + \ln y = \ln \left( {xy} \right)$, we get,
$\Rightarrow \log x\left( {x - 9} \right) = \log 10$,
Now applying exponents on both sides we get,
$\Rightarrow {e^{\log x\left( {x - 9} \right)}} = {e^{\log 10}}$,
Now we know that ${e^{\ln x}} = x$, we get,
$\Rightarrow x\left( {x - 9} \right) = 10$,
Now simplifying we get,
$\Rightarrow {x^2} - 9x = 10$,
Now factorising the equation, by adding $\dfrac{{81}}{4}$on both sides we get,
$\Rightarrow {x^2} - 9x + \dfrac{{81}}{4} = 10 + \dfrac{{81}}{4}$,
Now simplifying we get,
$\Rightarrow {x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{40 + 81}}{2}$,
Now write the left hand side as a perfect square, we get,
$\Rightarrow {\left( {x - \dfrac{9}{2}} \right)^2} = \dfrac{{121}}{4}$,
Now taking out the square root, we get,
$\Rightarrow x - \dfrac{9}{2} = \pm \dfrac{{11}}{2}$,
Now $x$ will have two values and they are, $x - \dfrac{9}{2} = \dfrac{{11}}{2}$and $x - \dfrac{9}{2} = - \dfrac{{11}}{2}$,
Taking the first value, we get,
$\Rightarrow x - \dfrac{9}{2} = \dfrac{{11}}{2}$,
Now take the L.C.M we get,
$\Rightarrow 2x - 9 = 11$,
Now add 9 on both sides we get,
$\Rightarrow 2x - 9 + 9 = 11 + 9$,
Now simplifying we get,
$\Rightarrow 2x = 20$,
Now divide both sides with 20, we get,
$\Rightarrow \dfrac{{2x}}{2} = \dfrac{{20}}{2}$,
Now simplifying we get,
$\Rightarrow x = 10$,
Now taking the second value, i.e.,
\Rightarrow $$$$x - \dfrac{9}{2} = - \dfrac{{11}}{2},
Now take the L.C.M we get,
$\Rightarrow 2x - 9 = - 11$,
Now add 9 on both sides we get,
$\Rightarrow 2x - 9 + 9 = - 11 + 9$,
Now simplifying we get,
$\Rightarrow 2x = - 2$,
Now divide both sides with 2, we get,
$\Rightarrow \dfrac{{2x}}{2} = \dfrac{{ - 2}}{2}$,
Now simplifying we get,
$\Rightarrow x = - 1$,
So, the values for $x$ are 10 and -1, but $x = - 1$ is impossible, because the logarithm of a negative number is undefined.
So, the value of $x$ is $10$.

$\therefore$ The value of the given function $\log x + \log \left( {x - 9} \right) = 1$ will be equal to $10$.

Note:
A logarithm is a mathematical operation that determines how many times a certain number, called the base, is multiplied by itself to reach another number, we know that logarithm is the power to which a number must be raised in order to get some other number, and the base unit is the number being raised to a power. In these types of questions, we use logarithmic properties and formulas, and some of useful formulas are:

1. ${\log _a}xy = {\log _a}x + {\log _a}y$
2. $\log x - \log y = \log \left( {\dfrac{x}{y}} \right)$
3. ${\log _a}{x^n} = n{\log _a}x$ ,
4. ${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$,
5. ${\log _{\dfrac{1}{a}}}b = - {\log _a}b$,
6. ${\log _a}a = 1$,
7. ${\log _{{a^x}}}b = \dfrac{1}{x}{\log _a}b$.