Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How does one solve ${\log _2}x = {\log _4}(x + 6)$ ?

Last updated date: 05th Mar 2024
Total views: 340.8k
Views today: 9.40k
Verified
340.8k+ views
Hint:For simplifying the original equation , firstly used logarithm property
${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$ then take base two exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the given equation .

Formula used:
We used logarithm properties i.e.,
${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$
And
${b^{{{\log }_b}a}} = a$ , the logarithm function says $\log x$ is only defined when $x$ is greater than zero.

Complete solution step by step:
It is given that ,
${\log _2}x = {\log _4}(x + 6)$ ,
We have to solve for $x$ .
Now using logarithm property ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$ ,
We will get,
${\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{{{{\log }_2}(4)}}$
Now , simplify the equation , we will get the following result ,
$\Rightarrow {\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{{{{\log }_2}{{(2)}^2}}}$ (using $\log {a^b} = b\log a$)
$\Rightarrow {\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{{2{{\log }_2}(2)}}$
Using ${\log _a}a = 1$ ,
$\Rightarrow {\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{2}$
$\Rightarrow 2{\log _2}x = {\log _2}(x + 6) \\ \Rightarrow {\log _2}{(x)^2} = {\log _2}(x + 6) \\$ (using $\log {a^b} = b\log a$)
Now , by assuming the base of the logarithm to be $2$ ,then take the base $2$ exponential of both sides of the equation, we will get the following result ,
For equation one ,
${2^{{{\log }_2}\left( {{x^2}} \right)}} = {2^{{{\log }_2}(x + 6)}}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get the following result ,
${x^2} = x + 6$
Or
${x^2} - x - 6 = 0$
$\Rightarrow {x^2} - 3x + 2x - 6 = 0 \\ \Rightarrow x(x - 3) + 2(x - 3) = 0 \\ \Rightarrow (x + 2)(x - 3) = 0 \\ \Rightarrow x = - 2 \\ and \\ x = 3 \\$
Now recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero.
Therefore, in our original equation ${\log _2}x = {\log _4}(x + 6)$ ,
Here,
$(x) > 0 \\ and \\ (x + 6) > 0 \\$ ,
When $x = - 2$ , it is less than zero.
Therefore , we reject $x = - 2$ .
And when $x = 3$ , it is greater than zero .
Therefore, we have our solutions i.e., $x = 3$ .

Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero. While performing logarithm properties we have to remember certain conditions , our end result must satisfy the domain of that logarithm .