$$(\log_{p^{1/n}} y)(\log_{y^{1/n}} p) = 16,$$
where the logarithms are taken to the bases $p^{1/n}$ and $y^{1/n}$.
The value of $n$ is ________
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.
Recall the logarithm property: $\log_{a^k} b = \frac{1}{k} \log_a b$. Applying this to our terms:
Substitute these simplified terms back into the original equation:
$(n \log_p y)(n \log_y p) = 16$ $n^2 (\log_p y)(\log_y p) = 16$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.
$n = 4$The value of $n$ that satisfies the given conditions is 4.
For positive non-zero real variables $p$ and $q$, if
$\log (p^2 + q^2) = \log p + \log q + 2 \log 3$,
then, the value of $\frac{p^4+q^4}{p^2q^2}$ 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
A petrified wood fossil was discovered with 8 g of $^{14}C$. The decay of $^{14}C$ over time is given by:
$N_T = N_0 e^{-0.0001216T}$
If the half-life of $^{14}C$ is 5700 years, and the fossil initially had 32 g of $^{14}C$, the age of the fossil in years is ______.