Parsing mathematical notation
I'm mentally dissecting some rendered math, and I see a small ? under what could be \(\sqrt{2}\), followed by a radical sign ending before 64. I think the notation is likely \(\log_{\sqrt{2}}64\), where the base is \(\sqrt{2}\) written lower than "log," and 64 sits at the normal baseline. In the image, I notice what seems like log, with \(\sqrt{2}\) beneath it, followed by 64. I might be spotting a pattern here!Examining logarithmic expression
I’m looking at the numerator in the image around line 7, which appears to be log_? 12^{?}. It could read as \(\frac{\log_4 12^{17}}{\log_4 12}\), where the numerator seems to be log_? 12^{17}. As I analyze the text more closely, it's clear that the numerator has the exponent 17. However, there's a possibility that it could also be \(\frac{\log_4 12^{12}}{\log_4 12}\). I'll double-check what's actually visible.Analyzing logarithmic expressions
The image shows 12^{?} with a superscript that looks like "17." The denominator seems to be log_4 12, suggesting it could be written as \( \frac{\log_4 12^{17}}{\log_4 12}\), which likely results in the answer being 17. I'm considering if the numerator could instead be \(\log_9 12^{17}\) over \(\log_9 12\), but that doesn't change the exponent outcome. The denominator is probably base 4, confirming the common logarithmic properties. Now, I’ll check the next expression, which is \(\log_{\sqrt{7}}8^{20}\).Clarifying logarithmic expressions
I think I might have misread something. It's likely that I'm dealing with \(\log_{\sqrt{8}} 8^{20}\), which returns 40. The subscript should be a radical with 8, not the square root of 7. In line 8, it seems to read \(\log_{\sqrt{8}} 8^{20}\). It’s clear to me now that the base is \(\sqrt{8}\) and the argument is \(8^{20}\).
For line 5, I might be looking at \(\log_{\sqrt{2}} 64\), which equals 12. I need to confirm whether the radical in the image refers to the argument or the base.Interpreting logarithmic expressions
I’m looking at the expression \(\log_{\sqrt{2}} 64\). The radical appears as a subscript close to the lower left of "log," while 64 is written normally. It seems like the base is indeed \(\sqrt{?}\).
If I'm considering \(\log_{2} \sqrt{64}\), the radical is higher before the 64, which gives a different visual. In line 5, I see something like “log” followed by a small radical to the right. I can focus on this by mentally zooming in like I would in a browser!Analyzing logarithmic expressions
I see a radical sign that might represent the square root of 64 with a log base of 2. If it's \(\log_2\sqrt{64}\), that makes sense, as the radical bar would extend over 64. I'm considering a sequence of log-related calculations, with values ranging from products to quotients. I think \(\log_2\sqrt{64}\) equals 3. I’ll explore if this sequence ties back to a worksheet or a broader educational context!Examining logarithmic expressions visually
I see a header that suggests this is about preparing for something related to exams in math. There’s a problem in line #5 that reads as log_{?}√64, which I suspect could be \(\log_2\sqrt{64}\). Let me double-check all the base and subscript sizes. Line 4 shows log_{81}(log_4 64). In line 5, I notice the small subscript might not have a radical after it. The visual suggests \(\log_2 \sqrt{64}\) could be the right interpretation, and I think I need to focus on that base!Evaluating math expressions
I see the number "17" raised above the normal "12," and it seems likely that's the point. For another part, if I take #5, it looks like I'm considering "log_2 √64," which I think should equal 3. I'm curious if I've worked that out correctly or if there's something I've overlooked. Let's verify that calculation, just to be sure!Calculating logarithm expression
I’m working with the expression \( \log_{\sqrt{8}} 8^{20} \). First, I recognize that \(\sqrt{8}\) is equivalent to \( 8^{1/2} \). Then, I use the change of base formula, leading to \( \log_{8^{1/2}} 8^{20} = \frac{20}{1/2} \), which gives me 40. It's important to calculate each step carefully, as this is a straightforward evaluation of logarithmic expressions.Analyzing log notation
I’m thinking about the line formula, possibly \(\log_{\sqrt{2}}64\). The visual would show log with a subscript √2, followed by 64. I notice that the radical's horizontal bar should be at subscript height, not covering 64. I'm wondering if there's a bar directly over 64. I see that the common notation in Russian print is \(\log_2 \sqrt{64}\), which looks interesting since it displays log with the subscript 2. This is a bit tricky!Considering logarithmic values
The problem asks for the log value, giving an integer result of 3. I'm looking at \(\log_{\sqrt{2}}64\), which evaluates to 12. It also tests the logarithm using a root base. I wonder which one is the likely choice for the question. It seems like there's some analysis to figure out which log base is more commonly selected or perhaps more intuitive for the problem at hand. There’s a lot to consider when comparing these values!Analyzing logarithmic expressions
I’m thinking about whether the problem could be \(\frac{\log_4 12^{17}}{\log_4 12}\). Then there's the question of if log_2 sqrt{64} needs parentheses for clarity, so it should be \(\log_2(\sqrt{64})\). I wonder if the expression can be simplified to give an answer of 11, perhaps related to \(7^{4-\log_7 4}\). I need to state this clearly before finalizing my response.
Краткие ответы
- \(6\)
- \(3\)
- \(4\)
- \(\dfrac14\)
- \(3\)
- \(4\)
- \(17\)
- \(40\)
- \(\dfrac{243}{2}\)
- \(\dfrac1{100}\)
- \(\dfrac{2401}{4}\)
- \(x=3\)
- \(x=\dfrac35\)
- \(x=-\dfrac{109}{50}\)
- \(x=5\)
- \(x=0\)
Решение
Используем свойства:
\[ \log_a b+\log_a c=\log_a(bc), \]
\[ \log_a b-\log_a c=\log_a\frac bc, \]
\[ \log_a b^n=n\log_a b, \qquad a^{\log_a b}=b. \]
\[ \log_3 145{,}8+\log_3 5 =\log_3(145{,}8\cdot5) =\log_3 729=6. \]
\[ \log_4 192-\log_4 3 =\log_4\frac{192}{3} =\log_4 64=3. \]
\[ \log_2 80+\log_2 0{,}2 =\log_2(80\cdot0{,}2) =\log_2 16=4. \]
\[ \log_4 64=3, \] поэтому \[ \log_{81}(\log_4 64)=\log_{81}3=\frac14, \] так как \(81=3^4\).
\[ \log_2\sqrt{64}=\log_2 8=\log_2 2^3=3. \]
\[ \log_7 49=2, \] \[ \log_2(\log_7 49+14) =\log_2(2+14) =\log_2 16=4. \]
\[ \frac{\log_4 12^{17}}{\log_4 12} = \frac{17\log_4 12}{\log_4 12} =17. \]
\[ \log_{\sqrt8}8^{20} = \frac{20}{1/2}=40, \] поскольку \(\sqrt8=8^{1/2}\).
\[ 9^{3-\log_9 6} =\frac{9^3}{9^{\log_9 6}} =\frac{729}{6} =\frac{243}{2}. \]
\[ 3^{-2\log_3 10} =\left(3^{\log_3 10}\right)^{-2} =10^{-2} =\frac1{100}. \]
\[ 7^{4-\log_7 4} =\frac{7^4}{7^{\log_7 4}} =\frac{2401}{4}. \]
\[ \log_4(11x-17)=2 \] \[ 11x-17=4^2=16, \] \[ 11x=33,\qquad x=3. \]
Проверка: \(11\cdot3-17=16>0\).
\[ \log_9(20x-9)=\log_9 3 \] \[ 20x-9=3, \] \[ 20x=12,\qquad x=\frac35. \]
\[ \log_6(-14x-1)=\log_6(6x+18)+1. \]
Так как \(1=\log_6 6\), получаем:
\[ \log_6(-14x-1) =\log_6\bigl(6(6x+18)\bigr). \]
Следовательно,
\[ -14x-1=36x+108, \]
\[ -50x=109, \qquad x=-\frac{109}{50}. \]
ОДЗ: \(-3<x<-\dfrac1{14}\), найденное значение подходит.
- \[ \log_{10}(2x-6)+\log_{10}2=\log_{10}8 \]
\[ \log_{10}\bigl(2(2x-6)\bigr)=\log_{10}8, \]
\[ 4x-12=8, \qquad x=5. \]
- \[ \log_{19}(4x+8)-\log_{19}4=\log_{19}2 \]
\[ \log_{19}\frac{4x+8}{4}=\log_{19}2, \]
\[ \frac{4x+8}{4}=2, \]
\[ 4x+8=8, \qquad x=0. \]
Проверка: при \(x=0\) получаем \(\log_{19}8-\log_{19}4=\log_{19}2\).