Let $a = 30!$, $b = 50!$, and $c = 100!$. Consider the following numbers: $log_a c$, $log_c a$, $log_b a$, $log_a b$ Which one of the following inequalities is CORRECT?
To solve this problem, we need to evaluate the given logarithmic expressions and determine which of the provided inequalities is correct. Let's analyze each expression:
We are given:
- \(a = 30!\), \(b = 50!\), \(c = 100!\)
- Expressions to evaluate: \(\log_a c\), \(\log_c a\), \(\log_b a\), \(\log_a b\)
Let's compute each logarithm using the property \(\log_x y = \frac{\log y}{\log x}\):
- \(\log_a c = \frac{\log(100!)}{\log(30!)}\)
Since \(100! > 30!\), this is a large number. - \(\log_c a = \frac{\log(30!)}{\log(100!)}\)
Since \(30! < 100!\), this is a small number. - \(\log_b a = \frac{\log(30!)}{\log(50!)}\)
Since \(30! < 50!\), this is a relatively small number. - \(\log_a b = \frac{\log(50!)}{\log(30!)}\)
Since \(50! > 30!\), this is a relatively large number.
Now, let's write out the approximate comparative sizes based on the factorial properties:
- \(\log_c a\) will be the smallest because \(\log(30!) \ll \log(100!)\).
- \(\log_b a\) will be larger than \(\log_c a\) but smaller than \(\log_a b\).
- \(\log_a b\) will be next since \(\log(50!) > \log(30!)\).
- \(\log_a c\) will be the largest because \(\log(100!) \gg \log(30!)\).
This order matches the first option provided:
\(\log_c a < \log_b a < \log_a b < \log_a c\)
Therefore, the correct inequality is:
\(\log_c a < \log_b a < \log_a b < \log_a c\)
Important Questions from Logarithms
- Real numbers $y$, $p$, and $n$ (all greater than 1) satisfy
$$(\log_{p^{1/n}} y)(\log_{y^{1/n}} p) = 16,$$
where the logarithms are taken to the bases $p^{1/n}$ and $y^{1/n}$.
The value of $n$ is ________ Consider two distinct positive real numbers $m, n$, with $m > n$.
Let $x = n^{\log_{10}(m)}$ and $y = m^{\log_{10}(n)}$. The relation between $x$ and $y$ is _______.
- If $\log_x (5/7) = -1/3$, then the value of $x$ is
- A value of x that satisfies the equation $ \log x + \log (x - 7) = \log (x + 11) + \log 2 $ is
For a real number $x > 1$,
$\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} = 1$
The value of $x$ is
