Question

How do you solve \[{\log _2}(x - 6) + {\log _2}(x - 4) = {\log _2}x\]?

Answer 1

Hint: We use the concept and properties of logarithm to solve this question. Since the base is the same on both sides of the equation and also between all the three terms, we use the property of adding log values on the left hand side and then take exponent function on both sides of the equation. Cancel log function by exponential function and equate the values on both sides. Solve the quadratic equation formed using any of the methods suitable: determinant or factorization.
* \[\log m + \log n = \log mn\]
* The general quadratic equation is \[a{x^2} + bx + c = 0\]
* For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step solution:
We have to solve the equation \[{\log _2}(x - 6) + {\log _2}(x - 4) = {\log _2}x\]
We are given all the three values of log with base 2.
We use the property of logarithm that states that \[\log m + \log n = \log mn\] to solve the left hand side of the equation (1).
\[ \Rightarrow {\log _2}\left[ {(x - 6)(x - 4)} \right] = {\log _2}x\]
Now we solve the value inside the bracket in left hand side of the equation
\[ \Rightarrow {\log _2}\left[ {(x(x - 4) - 6(x - 4)} \right] = {\log _2}x\]
Multiply the terms outside the bracket to the values inside the bracket in left hand side of the equation
\[ \Rightarrow {\log _2}\left[ {{x^2} - 4x - 6x + 24} \right] = {\log _2}x\]
Add like terms or the terms with the same variables in the left hand side of the equation.
\[ \Rightarrow {\log _2}\left[ {{x^2} - 10x + 24} \right] = {\log _2}x\]
Take exponential function on both sides of the equation
\[ \Rightarrow {e^{{{\log }_2}\left[ {{x^2} - 10x + 24} \right]}} = {e^{{{\log }_2}x}}\]
Cancel exponential function by log function on both sides of the equation
\[ \Rightarrow {x^2} - 10x + 24 = x\]
Shift all values to left hand side of the equation
\[ \Rightarrow {x^2} - 10x + 24 - x = 0\]
\[ \Rightarrow {x^2} - 11x + 24 = 0\]
Now we compare this quadratic equation with general quadratic equation i.e. \[a{x^2} + bx + c = 0\]
On comparing with general quadratic equation \[a{x^2} + bx + c = 0\], we get \[a = 1,b = - 11,c = 24\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow x = \dfrac{{ - ( - 11) \pm \sqrt {{{( - 11)}^2} - 4 \times 1 \times 24} }}{{2 \times 1}}\]
Square the values inside the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{11 \pm \sqrt {121 - 96} }}{4}\]
\[ \Rightarrow x = \dfrac{{11 \pm \sqrt {25} }}{4}\]
Cancel square root by square power in the numerator
\[ \Rightarrow x = \dfrac{{11 \pm 5}}{4}\]
So, \[x = \dfrac{{11 + 5}}{4}\] and \[x = \dfrac{{11 - 5}}{4}\]
I.e. \[x = \dfrac{{16}}{4}\] and \[x = \dfrac{6}{4}\]
Cancel possible factors from numerator and denominator
\[x = 4\] and \[x = \dfrac{3}{2}\]
i.e. \[x = 4\] and \[x = 1.5\]

\[\therefore \]Solution of the equation \[{\log _2}(x - 6) + {\log _2}(x - 4) = {\log _2}x\] is \[x = 4\] and \[x = 1.5\].

Note: Many students make mistake of opening the values using log base on both sides as they try to apply the formula of difference of log into division which cannot be applied here as difference is between the terms of same log function like \[{\log _2}(x - 6)\], it cannot be written as \[\dfrac{{{{\log }_2}x}}{{{{\log }_2}6}}\]. Keep in mind for this to be possible, we should have the function \[{\log _2}x - {\log _2}6x\].