$$\log_2 x + \log_{\sqrt{2}} x = 48,$$
then the value of $x$ is _________________
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 $2^{16}$.
Important Questions from Logarithms
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 $\frac{p^4+q^4}{p^2q^2}$ isFor a real number $x > 1$,
$\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} = 1$
The value of $x$ isA 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 ______.
- For integers $a$, $b$ and $c$, what would be the minimum and maximum values respectively of $a + b + c$ if $\log |a| + \log |b| + \log |c| = 0$?
- The value of the expression $\frac{1}{1+\log_u vw} + \frac{1}{1+\log_v wu} + \frac{1}{1+\log_w uv}$ is
