Question
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
The correct answer is
24
Logarithm Equation Solving
The problem requires finding the value of $x$ for the equation:
$ \frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} = 1 $
We are given the condition $x > 1$. The solution involves applying key logarithm properties.
Logarithm Properties Applied
- Change of Base Formula: $ \frac{1}{\log_a b} = \log_b a $. This allows us to change the base of the logarithm to $x$.
- Sum Rule for Logarithms: $ \log_b m + \log_b n = \log_b (mn) $. This helps combine multiple logarithm terms with the same base.
Solution Steps
- Transform the equation: Using the change of base formula, rewrite each term with base $x$: $ \log_x 2 + \log_x 3 + \log_x 4 = 1 $
- Combine logarithmic terms: Apply the sum rule for logarithms: $ \log_x (2 \times 3 \times 4) = 1 $ Simplify the product inside the logarithm: $ \log_x 24 = 1 $
- Convert to exponential form: Recall that $ \log_b a = c $ is equivalent to $ b^c = a $. Applying this definition: $ x^1 = 24 $
- Determine the value of x: From the exponential form, we find: $ x = 24 $ The obtained value $x=24$ satisfies the initial condition $x > 1$.
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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
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?
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