Question

How do you condense $6\log (x) - \log (17)$ to a simple logarithm?

Answer 1

Hint: Check out all the laws of logarithms. The most important law of logarithms used here is the powers of logarithms and the subtraction of logarithms. First, convert the first term into power form and then use the subtraction law to subtract both the terms to get a single term.

Formulas:
Law of powers of logarithms, if we have the function, $f(x) = {\log _a}({b^c})$ .Then we can convert into power form as, $f(x) = c{\log _a}(b)$ .
Law of subtraction of logarithms, if we have the expression, ${\log _a}(b) - {\log _a}(c)$ . It is equal to ${\log _a}(\dfrac{b}{c})$.

Complete step-by-step answer:
Assuming that the bases of the logarithmic terms are the same,
the given expression is $6\log (x) - \log (17)$.
Now, consider the first term which is $6\log (x)$.
Since we know that $\log ({b^c}) = c\log (b)$,
Compare the RHS of the formula with the term.
$ \Rightarrow $$c = 6;b = x$
Now, write it in the power form.
$ \Rightarrow \log ({x^6})$
Now, rewrite all the terms together.
$ \Rightarrow \log ({x^6}) - \log (17)$
By using the subtraction law of logarithms which is, $\log (b) - \log (c) = \log (\dfrac{b}{c})$
$ \Rightarrow b = {x^6};c = 17$
On substitution we get,
$ \Rightarrow \log (\dfrac{{{x^6}}}{{17}})$
$\therefore 6\log (x) - \log (17)$ on simplifying we get $\log (\dfrac{{{x^6}}}{{17}})$.

Additional Information: The logarithm of a given constant $y$ is the exponent to which another fixed constant, the base $b$ , must be raised, to produce that constant $y$.
$ \Rightarrow {\log _b}({b^x}) = x$
Sometimes logarithm is written without a base. In this case, we must assume that the base is $10$. It is called the “common logarithm”.

Note:
One should ensure that the base of the given logarithms is the same before evaluating the expression using the laws of the logarithms. While using the subtraction law of logarithms ensure which term is written in the numerator and which in the denominator. Always the first term will be in the numerator whereas the second term in the denominator.