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How does one solve ${\log _{27}}x = 1 - {\log _{27}}(x - 0.4)$ ?

seo-qna
Last updated date: 13th Jun 2024
Total views: 372.9k
Views today: 4.72k
Answer
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372.9k+ views
Hint:For simplifying the original equation , firstly used logarithm property
$\log a + \log b = \log ab$ then take base twenty seven 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.
$\log a + \log b = \log (ab)$ ,
${b^{{{\log }_b}a}} = a$
and
We also used quadratic formula i.e.,
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete solution step by step:
It is given that ,
${\log _{27}}x = 1 - {\log _{27}}(x - 0.4)$ ,
We have to solve for $x$ .
We can manipulate the above equation as ,
${\log _{27}}x + {\log _{27}}(x - 0.4) = 1$
Now using logarithm property $\log a + \log b = \log ab$ ,
We will get,
${\log _{27}}x(x - 0.4) = 1$
Now , by assuming the base of the logarithm to be $27$ ,then take the base $27$ exponential of both sides of the equation, we will get the following result ,
For equation one ,
${27^{{{\log }_{27}}\left( {x(x - 0.4} \right)}} = {27^1}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get the following result ,
$x(x - 0.4) = 27$
Or
$
{x^2} - 0.4x - 27 = 0 \\
or \\
10{x^2} - 4x - 270 = 0 \\
$
For finding roots of the original equation, we have to use quadratic formula i.e.,
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now identify $a,b,c$ from the original equation given below,
$ \Rightarrow 10{x^2} - 4x - 270 = 0$
$
\Rightarrow a = 10 \\
\Rightarrow b = - 4 \\
\Rightarrow c = - 270 \\
$
Put these values into the formula of finding the roots of quadratic equations,
$x = \dfrac{{4 \pm \sqrt {{4^2} - 4*10*( - 270)} }}{{2*10}}$
After simplifying and by evaluating exponents and square root of the above equation we get the following simplified expression,
$x = \dfrac{{4 \pm 104}}{{20}}$
To find the roots of the equations , separate the particular equation into its corresponding parts : one part with the plus sign and the other with the minus sign ,
$
\Rightarrow {x_1} = \dfrac{{4 + 104}}{{20}} \\
and \\
\Rightarrow {x_2} = \dfrac{{4 - 104}}{{20}} \\
$
Simplify and then isolate $x$ to find its corresponding solutions!
$
\Rightarrow {x_1} = 5.4 \\
and \\
\Rightarrow {x_2} = - 5 \\
$
Now recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero.
Therefore, in our original equation ${\log _{27}}x = 1 - {\log _{27}}(x - 0.4)$ ,
Here,
$
(x - 0.4) > 0 \\
and \\
(x) > 0 \\
$
Now after evaluating both the values, $ - 5$ is rejected because it is less than zero. While $5.4$ is greater than zero and $5.4 - 0.4 > 0$ .
Therefore, we have our solution i.e., $5.4$ .

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 .