Question
If a positive real $x$ satisfies the following equation
$$\log_2 x + \log_{\sqrt{2}} x = 48,$$
then the value of $x$ is _________________
The correct answer is
$2^{16}$
Solving the Logarithm Equation $\log_2 x + \log_{\sqrt{2}} x = 48$
Key Logarithm Properties Used
- Change of base property: $\log_{b^k} a = \frac{1}{k} \log_b a$.
- Combining like terms.
Step-by-Step Solution
- Simplify the second term: The base of the second logarithm is $\sqrt{2}$, which can be written as $2^{1/2}$. Using the property $\log_{b^k} a = \frac{1}{k} \log_b a$, we have:
$\log_{\sqrt{2}} x = \log_{2^{1/2}} x = \frac{1}{1/2} \log_2 x = 2 \log_2 x$.
- Substitute back into the equation: Replace $\log_{\sqrt{2}} x$ with $2 \log_2 x$ in the original equation:
$\log_2 x + 2 \log_2 x = 48$.
- Combine logarithmic terms: Add the terms involving $\log_2 x$:
$3 \log_2 x = 48$.
- Isolate the logarithm: Divide both sides by 3:
$\log_2 x = \frac{48}{3}$
$\log_2 x = 16$.
- Solve for $x$: Convert the logarithmic equation to its exponential form. If $\log_b y = c$, then $y = b^c$.
$x = 2^{16}$.
Final Answer
The value of $x$ that satisfies the equation $\log_2 x + \log_{\sqrt{2}} x = 48$ is $2^{16}$.
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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
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
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