# Tag Info

6

should be $$\left( \log _{ 3 }{ x } -1 \right) \left( \log _{ 3 }{ x } -2 \right) =0$$ $$\log _{ 3 }{ x } =1\Rightarrow \quad x=3\\ \log _{ 3 }{ x } =2\Rightarrow x=9$$

5

$$\frac{\log25}{\log5}+\frac{\log625}{\log5}=\frac{\log5^2}{\log5}+\frac{\log5^4}{\log5}=\frac{2\log5}{\log5}+\frac{4\log5}{\log5}=2+4=6$$

5

$2 = \frac{1}{2}^{-1}$ $\sqrt[3]{4} = 2 ^{2/3}= (1/2)^{-2/3}$ $\log_{\frac{1}{2}}{(2\sqrt[3]{4})} = -1 - 2/3 = -5/3$

4

The natural logarithm is continuous on its domain, so you can do that, but I don't recommend it. Another approach is to rewrite the natural logarithm as $$\ln(n^n\cdot(n+1)^{-n-1})=n\ln(n)-(n+1)\ln(n+1).$$ Then the result follows from the fact that the natural logarithm increases strictly to infinity.

3

Note that $\log_b a^n = n \log_b a$, $\log a + \log b = \log ab$ Rewrite in terms of powers of 5: $$\frac{ \log 5^2 + \log 5^4}{\log 5}= \frac{\log (5^2 \times 5^4)}{\log 5} = \frac{6\log 5}{\log 5} = ?$$

3

\begin{align*} \log_{1/2}(2\sqrt[3]4) & =\log_{1/2}(2)+\frac{1}3\log_{1/2}(4) \\ & =-1-\frac{2}3 \end{align*}

3

It's hard to say there's "a" correct definition. But what many texts do is not define $\ln$ as the inverse of $\exp$, but define each individually then show their relation. In particular, $\ln x = \int_{1}^{x} \frac{1}{t} \mathrm{d}t$ makes sense as it's the unique function with derivative $1/x$ and with value $0$ at $1$. From this, you can show $\ln (a b) = ... 3 $$x^{ log_{ 2 }x }+\frac { 16 }{ x^{ log_{ 2 }x } } =17\\ { x }^{ 2\log _{ 2 }{ x } }-17x^{ log_{ 2 }x }+16=0\\ \left( x^{ log_{ 2 }x }-16 \right) \left( x^{ log_{ 2 }x }-1 \right) =0\\ x^{ log_{ 2 }x }=16\Rightarrow \log _{ 2 }{ x^{ log_{ 2 }x } } =\log _{ 2 }{ 16 } \Rightarrow { \left( \log _{ 2 }{ x } \right) }^{ 2 }=4\Rightarrow \log _{ 2 }{ x } =\pm ... 3 hint: that means 2^n \geq n and it can be done with easy induction 2 Other answers have offered the solution as expected. I wanted to highlight that such sort of questions tend to encourage more hand waving than doing any service to mathematics. When you write$$\log (1 + x) \approx A + Bx + Cx^{2}\tag{1}$$you actually mean that there is an error R(x) involved so that$$\log(1 + x) = A + Bx + Cx^{2} + R(x)\tag{2}$$for ... 2 let$$y=x^{log_2x}$$your equation becomes,$$y+\frac{16}y=17$$solve it, you get two solutions: 1 and 16. Now it becomes less horrible,$$x^{log_2x}=1 ~or~ 16 $$This leads to solution x =1, 2^2 and 2^{-2} 2 You cannot ignore the exponent or the multiplication. In this case, set temporarily t=\log_3x so the equation becomes$$ t^2-3t+2=0 $$which has roots t=1 and t=2. Thus you get the two equations$$ \log_3x=1 $$and$$ \log_3x=2 $$Can you finish them? Your method would be sound, too, provided you did the decomposition right. 1 Hint. One may write$$ x=\log_{2a}a=\frac{\log(a)}{\log(2a)},\quad y=\log_{3a}(2a)=\frac{\log(2a)}{\log(3a)},\quad z=\log_{4a}(3a)=\frac{\log(3a)}{\log(4a)} $$giving$$ xyz-2yz=\frac{\log(a)}{\log(2a)}\cdot\frac{\log(2a)}{\log(3a)}\cdot\frac{\log(3a)}{\log(4a)}-2\cdot\frac{\log(2a)}{\log(3a)}\cdot\frac{\log(3a)}{\log(4a)} $$that is$$ xyz-2yz=\frac{\... 1 Let$ \frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k\rightarrow a=e^{k(b-c)}, b=e^{k(c-a)},c=e^{k(a-b)}$. Then the required value is$ e^{k(b^2-c^2)}. e^{k(c^2-a^2)}. e^{k(a^2-b^2)}=1$1$\textbf{Hint:}$Try proving that if$f$is nowhere zero on a simply connected domain$U$, then there exists a holomorphic branch of$\text{log}~f$. To prove this, you'll need to use the fact that a holomorphic branch of$\text{log}~f$exists if and only if$\dfrac{f'}{f}$has a primitive in$U$. 1 $$\lim_{n \to \infty}\frac{n^n}{(n+1)^{n+1}} = \lim_{n \to \infty} \frac{1}{(1+\frac 1n)^n \cdot (n+1)} = 0$$ Now just take the logarithm and use the fact that the natural logarithm is a continuous function. 1 Yes, that works. The inner limit is at the edge of the domain for$\log$however. 1 As said in comments, I do not think that there is any hope for an analytical solution and numerical methods need to be considered. Using natural logarithms, the equation write $$\frac{\log (x+1)}{\log (5)}-\frac{\log (x-2)}{\log (4)}=1$$ So, let us consider the function $$f(x)=\frac{\log (x+1)}{\log (5)}-\frac{\log (x-2)}{\log (4)}-1$$ By inspection,$f(3)=\...

1

Calculators don't actually need to define logarithms -- it is enough that they can somehow calculate a number that is within a desired tolerance from the true value of the logarithm, and if a method achieves that, then it doesn't matter whether it's based on a principled definition of the logarithm function or not. Calculators that are based on binary ...

1

First, we have $$e^{\textstyle 10^{10^{2.8}}} = 10^{\textstyle \log_{10}(e) \cdot 10^{10^{2.8}}} = 10^{\textstyle 10^{\left( 10^{2.8} + \log_{10}\log_{10}(e) \right)}} = 10^{\textstyle 10^{10^{A}}}$$ where $$A = \log_{10}(10^{2.8} + \log_{10}\log_{10}(e)) \approx \log_{10}(630.96 - 0.36) \approx 2.7998$$ Now, setting $B=10^{10^A}$, do the same ...

1

In general, not necessarily. The left-hand side needs to be big enough. The smallest we can get is if we let $x \to -\infty$, which gives us $$10^{10^{10^{10^{x}}}} \to 10^{10^{10^{10^{-\infty}}}} \to 10^{10^{10^{0}}}=10^{10^{1}} = 10^{10}$$ so basically, we cannot reach down to $10^{10}$, but any number above that is large enough that we can find an $x$ ...

1

By definition of the prime counting function we have $\pi(p_n)=n$ for the $n$-th prime $p_n$. So, for $n\to \infty$, $$n=\pi(p_n)\sim \frac{p_n}{\log(p_n)},$$ which says $p_n\sim n\log(n)$.

1

As given in Olba12's answer, there is no closed form for the solution of equation $$\gamma = \ln\left(\frac{1}{\gamma} + 1\right)$$ (I checked using inverse symbolic calculators) and numerical methods are required. Considering $$f(\gamma)=\gamma - \ln\left(\frac{1}{\gamma} + 1\right)$$ $$f'(\gamma)=1+\frac{1}{\gamma (\gamma +1)}$$ and using (as the ...

1

From your comment you stated the following $$a_{n+1} = \ln \left( \frac{1}{a_n} + 1 \right)$$ Then denote the limit of $a_n$ to be $\gamma$ when $n \to \infty$. By definition it follows that $\lim_{n\to \infty} a_n = \lim_{n \to \infty} a_{n+1}$ hence. $$\gamma = \ln\left(\frac{1}{\gamma} + 1\right)$$ Which gives $$\gamma_1 = 0.... 1 Given a probability distribution p with n distinct outcomes x_i, the quantity$$\frac{H}{H_\text{max}} = \frac{-\sum_{i=1}^n p(x_i) \log(p(x_i))}{-\sum_{i=1}^n \frac{1}{n} \log(\frac{1}{n})} = -\sum_{i=1}^n \frac{p(x_i) \log(p(x_i))}{\log(n)}is sometimes called as the efficiency, or the normalized entropy. Wikipedia has a short paragraph on this, ... 1 First of all, I would use log base 2 instead of natural log because it's easier to talk about its meaning as the number of yes/no questions on average to guess the value. Given 20 choices, the maximum entropy distribution has entropy of 4.322 bits. While your distribution has 3.607 bits, which is 83% of the maximum possible value. Of course you can ... 1 @Michael Hardy offered a clear solution to the first part. Second Part solution The second part of the problem can be solved by applying the solution from the first part: \begin{align*} \left(1+\frac{1}{2n}\right)^{n+3} &< \left(1+\frac{1}{n}\right)^{n-1} \\ \ln \left(1+\frac{1}{2n}\right)^{n+3} &< \ln \left(1+\frac{1}{n}\right)^{n-1} &\... 1 First part you can solve it. So we can write according to question ln(1+x)=x-\frac{x^2}{2}. Now for the inequality, take ln both the sides and use above formluae. After appying above equation, you will get the inequality like this\frac{(n+3)(4n-1)}{4}<(n-1)(2n-1)$$which will turn into$$4n^2-23n+7>0$$which gives$$n\ge6 which is the final ...

Only top voted, non community-wiki answers of a minimum length are eligible