Question

How does one solve ${\log _{x - 2}}(9) = 2$ ?

Answer 1

Hint: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.

Formula used:
We used logarithm formula i.e.,
\[{\log _b}\left( x \right) = y\]
$ \Rightarrow {b^y} = x$
Where $x$ and $b$ are positive real numbers and $b$ is not equal to one.

Complete step by step answer:
We have the following equations ,
${\log _{x - 2}}(9) = 2$ ,
We can write this equation in exponential form using the formula of logarithm i.e.,
\[{\log _b}\left( x \right) = y\]
$ \Rightarrow {b^y} = x$
Where $x$and $b$are positive real numbers and $b$ is not equal to one.
By using the above formula, we get ,
$ \Rightarrow {(x - 2)^2} = 9$
For simplification we can write
$ \Rightarrow (x - 2) = \pm 3$
Now we can separate the above equation into two parts one with positive sign and other with negative sign as,
$(x - 2) = - 3 \\
\Rightarrow x = - 3 + 2 \\
\Rightarrow x = - 1 \\
(x - 2) = 3 \\
\Rightarrow x = 3 + 2 \\
\Rightarrow x = 5 \\ $
Now we have two values, but these values must satisfy the logarithm rule i.e.,
\[{\log _b}\left( x \right) = y\]
This implies this ${b^y} = x$ only when $x$and $b$are positive real numbers and $b$ is not equal to one.Therefore, for the given equation , ${\log _{x - 2}}(9) = 2$
$x - 2 > 0 \Rightarrow x > 2$
Therefore $x = - 1$ is rejected.

Hence, we have only one solution i.e., $x = 5$.

Additional Information:
Let we have a variable $a$ which is greater than zero for this particular section. Now,
\[lo{g_a}{\text{ }} = {\text{ }}0{\text{ }}and{\text{ }}lo{g_a}{\text{ }}a{\text{ }} = {\text{ }}1\]
Since we know that
\[{\log _b}\left( x \right) = y\]
$ \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:
\[{\log _b}\left( x \right) = y\]
$ \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.