$$(\log_{p^{1/n}} y)(\log_{y^{1/n}} p) = 16,$$
where the logarithms are taken to the bases $p^{1/n}$ and $y^{1/n}$.
The value of $n$ is ________
Solving the Logarithm Equation
We are given the equation involving real numbers $y, p, n$ (all greater than 1):
$(\log_{p^{1/n}} y)(\log_{y^{1/n}} p) = 16$We need to find the value of $n$. Let's simplify the logarithmic terms using logarithm properties.
Logarithm Property Application
Recall the logarithm property: $\log_{a^k} b = \frac{1}{k} \log_a b$. Applying this to our terms:
- The first term: $\log_{p^{1/n}} y = \frac{1}{1/n} \log_p y = n \log_p y$
- The second term: $\log_{y^{1/n}} p = \frac{1}{1/n} \log_y p = n \log_y p$
Substitute these simplified terms back into the original equation:
$(n \log_p y)(n \log_y p) = 16$ $n^2 (\log_p y)(\log_y p) = 16$Simplifying to Find n
Now, use the change of base property for logarithms: $\log_a b = \frac{1}{\log_b a}$. This means $\log_y p = \frac{1}{\log_p y}$. Substitute this into the equation:
$n^2 (\log_p y) \left(\frac{1}{\log_p y}\right) = 16$The term $(\log_p y)$ cancels out (since $y > 1$, $\log_p y$ is non-zero):
$n^2 = 16$To find $n$, we take the square root of both sides:
$n = \pm \sqrt{16}$ $n = \pm 4$The problem states that $n > 1$. Therefore, we choose the positive value.
$n = 4$Final Answer Determination
The value of $n$ that satisfies the given conditions is 4.
Important Questions from Logarithms
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 $x$ isLet $a = 30!$, $b = 50!$, 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?
