
What is ${\log _a}(\dfrac{1}{a})$?
Answer
481.2k+ views
Hint: A quantity representing the power to which a fixed number (the base) must be raised to produce a given number. A common and general logarithm is one with base 10, in such cases we don’t have to mention the value of the base. We use certain basic properties of logarithm such as the exponential property which states that:
${\log _b}{m^n} = n{\log _b}m$
This property holds true if and only if $m > 0$.(domain of a logarithmic function)
Complete step-by-step answer:
The given question is: ${\log _a}(\dfrac{1}{a})$
We can write $\dfrac{1}{a} = {a^{ - 1}}$
Now substituting this into the question, we get:
based on the exponential property ${\log _b}{m^n} = n{\log _b}m$
${\log _a}{a^{ - 1}} = - 1({\log _a}a)$
We know the basic property of a logarithmic function is that; when the base value is equal to the
the number whose logarithm we are finding, it is equal to 1.
Which implies: ${\log _a}a = 1$
Therefore, from the above results and equations we can tell that:
$( - 1) \times (1) = - 1$
The above synthesis is valid if and only if $a > 0$
Note: Logarithms are inverse of exponentials. Logarithm is the exponent or power to which a base must be raised to yield a given number. Logarithms are a convenient way to express large numbers. Operations like multiplication and division become addition and subtraction respectively. We need to remember basic properties of logarithms that make simplification easier.
${\log _b}{m^n} = n{\log _b}m$
This property holds true if and only if $m > 0$.(domain of a logarithmic function)
Complete step-by-step answer:
The given question is: ${\log _a}(\dfrac{1}{a})$
We can write $\dfrac{1}{a} = {a^{ - 1}}$
Now substituting this into the question, we get:
based on the exponential property ${\log _b}{m^n} = n{\log _b}m$
${\log _a}{a^{ - 1}} = - 1({\log _a}a)$
We know the basic property of a logarithmic function is that; when the base value is equal to the
the number whose logarithm we are finding, it is equal to 1.
Which implies: ${\log _a}a = 1$
Therefore, from the above results and equations we can tell that:
$( - 1) \times (1) = - 1$
The above synthesis is valid if and only if $a > 0$
Note: Logarithms are inverse of exponentials. Logarithm is the exponent or power to which a base must be raised to yield a given number. Logarithms are a convenient way to express large numbers. Operations like multiplication and division become addition and subtraction respectively. We need to remember basic properties of logarithms that make simplification easier.
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