Question

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

Answer 1

Hint:
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) \[{\log _{{a^x}}}b = \dfrac{1}{x}{\log _a}b\].