Question

How does one solve ${\log _x}6 = 0.5$ ?

Answer 1

You can solve it by simply writing it in exponential form using logarithm properties then reduce the exponent and lastly evaluate the values by checking whether they satisfy the definition of logarithm or not.

Complete solution step by step:
We have the following equations ,
${\log _x}6 = 0.5$ ,
Or we can also write it as ,
${\log _x}6 = \dfrac{1}{2}$
We can write this equation in exponential form using the formula of logarithm i.e.,
${\log _b}(x) = y \\\ \Rightarrow {b^y} = x \\\$
Where $x$ and $b$ are positive real numbers and $b$ is not equal to one.
Now compare the given equation with the general equation . we get the following result: the base of the given equation is equal to $x$ and this base must be a positive real number and must not equal to one also.
By using the above formula, we get ,
${(x)^{\dfrac{1}{2}}} = 6$
For simplification , squaring both sides .
We will get ,
$\Rightarrow {\left( {\sqrt x } \right)^2} = {6^2} \\\ \Rightarrow x = 36 \\\$
Here $x$ is a positive real number and not equals to one .
Therefore , we get the result that is $x = 36$ .
Formula used:
We used logarithm formula i.e.,
${\log _b}(x) = y \\\ \Rightarrow {b^y} = x \\\$
Where $x$ and $b$ are positive real numbers and $b$ is not equal to one.
Additional Information: Let we have a variable $a$ which is greater than zero for this particular section. Now,
${\log _a} = 0 \\\ and \\\ {\log _a}a = 1 \\\$
Since we know that
${\log _b}(x) = y \\\ \Rightarrow {b^y} = x \\\$
Where $x$ and $b$ are positive real numbers and $b$ is not equal to one.

Note: The relationship we used in this question is between logarithms and powers . This relationship is connecting exponents and logarithm as follow:
${\log _b}(x) = y \\\ \Rightarrow {b^y} = x \\\$
Where $x$ and $b$ are positive real numbers and $b$ is not equal to one. These are some necessary conditions for defining a logarithm function.