Question

How do you solve for \[{\log _7}\left( {\dfrac{1}{{49}}} \right) = x\]?

Answer 1

Hint: Logarithm is the inverse function of exponentiation. It means that the logarithm of a given number \[y\] is the exponent to which another number, the base \[b\], must be raised to produce that number \[y\]. To solve this question, we will use the definition of logarithm. We will put \[x\] in the exponential of \[7\] and equate it to \[\dfrac{1}{{49}}\]. Then from this equation we will find the value of \[x\].

Complete step by step answer:
We have;
\[{\log _7}\left( {\dfrac{1}{{49}}} \right) = x\]
From the definition of logarithm, we can write;
\[ \Rightarrow {7^x} = \dfrac{1}{{49}}\]
Writing the RHS in terms of \[7\]. We get;
\[ \Rightarrow {7^x} = {7^{ - 2}}\]
Now since the base is the same on both sides, we can equate the power. So, we have;
\[ \Rightarrow x = - 2\]

Note:
We can also solve this question by another method.
We have;
\[{\log _7}\left( {\dfrac{1}{{49}}} \right) = x\]
Writing \[\dfrac{1}{{49}} = {7^{ - 2}}\], we get;
\[ \Rightarrow {\log _7}{7^{ - 2}} = x\]
Now we will use the property mentioned in the additional details that; \[{\log _b}{x^n} = n{\log _b}x\]
So, we have;
\[ \Rightarrow - 2{\log _7}7 = x\]
Now we know that when the base and the argument of the logarithm is same then the logarithm is equal to one. So, we get,
\[ \Rightarrow - 2 \times 1 = x\]
So, we get;
\[ \Rightarrow x = - 2\]