Simplifying the Logarithmic Expression
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:
- Term 1 Denominator: $1+\log_u vw = \log_u u + \log_u vw = \log_u (u \cdot vw) = \log_u (uvw)$
- Term 2 Denominator: $1+\log_v wu = \log_v v + \log_v wu = \log_v (v \cdot wu) = \log_v (uvw)$
- Term 3 Denominator: $1+\log_w uv = \log_w w + \log_w uv = \log_w (w \cdot uv) = \log_w (uvw)$
Now, substitute these back into the expression:
$ E = \frac{1}{\log_u (uvw)} + \frac{1}{\log_v (uvw)} + \frac{1}{\log_w (uvw)} $
Applying Change of Base Property
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 $
Combining Logarithms and Final Value
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.
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
