The problem asks for the value of the expression:
$ E = \frac{1}{1+\log_u vw} + \frac{1}{1+\log_v wu} + \frac{1}{1+\log_w uv} $
We can simplify each term using logarithm properties. Recall that $1 = \log_b b$ for any base $b$. Let's rewrite the denominators:
Now, substitute these back into the expression:
$ E = \frac{1}{\log_u (uvw)} + \frac{1}{\log_v (uvw)} + \frac{1}{\log_w (uvw)} $
Using the logarithm property $\frac{1}{\log_a b} = \log_b a$, we can rewrite the expression:
$ E = \log_{uvw} u + \log_{uvw} v + \log_{uvw} w $
Using the logarithm property $\log_b x + \log_b y + \log_b z = \log_b (xyz)$, we combine the terms:
$ E = \log_{uvw} (u \cdot v \cdot w) $
$ E = \log_{uvw} (uvw) $
Finally, using the property $\log_b b = 1$:
$ E = 1 $
Therefore, the value of the expression is 1.
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 ______.