Hot answers tagged

15

Suppose $\log_2 7 = {a\over b}$ for two positive integers $a$ and $b$. $\log_2 7 = { \ln 7 \over \ln 2} = {a \over b}$ Cross multiply, $b \ln 7 = a \ln 2 \implies \ln ( 7^b ) = \ln (2^a)$ Take the $e^{( \ \ )}$ of both sides, $7^b = 2^a$ This is impossible for integers $a$ and $b$ because $7^b$ is always going to be an odd number, while $2^a$ will ...


6

Google's calculator is treating it as $8\times \log_{10} 2$ and the "log calculator" to which you link is finding $\log_2 8$. From the fact that $2^3=8$, it follows that $\log_2 8=3$. $8\log 2$ means $8\log_b 2$, where $b$ is some positive number other than $1$. The expression does not specify which number $b$ is. In some contexts in science and ...


6

I would think $8$ log $2$ means the person is usually looking for $8\log 2$ Even so, It's not very clear base you are in. Usually if you are in a science class Log is specifically referring the logarithm of a number base $10$. But mathematicians like to use Log to mean the logarithm base $e$. What you written is wrong, but let's just do the math behind it: ...


5

One has $$\log x = \int_1^x \frac{dt}{t} \leq \int_1^x dt = x-1 < x$$ Since $x\geq 1$, dividing by $x^2$ preserves the inequality.


4

Use the fact that $\log_a(b)=\frac{\log_c(b)}{\log_c(a)}$. This gives $$\log_2(x)\log_4(x)\log_8(x) = \log_2(x)\frac{\log_2(x)}{\log_2(4)}\frac{\log_2(x)}{\log_2(8)}=\frac16(\log_2(x))^3$$ So $\frac16(\log_2(x))^3=4.5$, hence $(\log_2(x))^3=27$, hence $\log_2(x)=3$, and thus $x=8$. It can be seen that $x=8$ indeed satisfies the given equation.


4

Here's one approach: Note: $\log_a a = 1$, $\log_b a = \frac{\log_n a}{\log_n b}$ $$ \log_2 x\times \log_4 x \times \log_8 x = 4.5 = \frac{9}{2}$$ change all logs to base 2: $$\log_2 x \times \frac{\log_2 x}{\log_2 4}\times \frac{\log_2 x}{\log_2 8} = \frac{9}{2} $$ $$ \log_2 x \times \frac{1}{2}\log_2 x \times \frac{1}{3}\log_2 x = \frac{9}{2}$$ ...


4

Note that $a+1\lt3^a$ if $a\gt1$ and $3(b+1)\lt2^b$ if $b\gt3$. Consequently $3(a+1)(b+1)\lt3^a\cdot2^b$ unless either $a=1$ or $0\le b\le3$. (Note, $a$ cannot be $0$, since the left hand side is divisible by $3$.) Thus we have five subcases to consider: $a=1$ and $6(b+1)=3\cdot2^b$ $b=0$ and $3(a+1)=3^a$ $b=1$ and $6(a+1)=3^a\cdot2$ $b=2$ and ...


4

If someone says "eight log $2$" they mean this: $$8\log2$$ Where $\log=\log_{10}$ and is interchanged with the terms "common log" or "log base $10$". It can be rewritten as $$8\log_{10}2$$ Edit: Though from my experience it almost always refers to base $10$, judging by the comments below this post "eight log 2" could also be rewritten as the following as ...


3

The solutions here are very nice. I will consider this problem as a number theory problem. To do this, I will need to adopt Case 2 and Case 3 from @Lanier Freeman 's solution. So, if we consider Case I only, we see that the number $n=3^a 2^b$ with $a, b\geq 0$ satisfies $$ 3\tau(n) = n $$ where $\tau(n)$ is the number of divisors of $n$. By an elementary ...


3

I've been playing around with this problem for half an hour now and I don't believe this is solvable through elementary logarithmic means. Your equation is equivalent to this: $$(a+1)(b+1)=3^{a-1}\cdot2^b$$ Case 1: Quadrant I & Axes Now consider the following for nonnegative integer solutions: $$3^{a-1}=a+1$$ $$2^b=b+1$$ The first equation is ...


3

The first point is: when you use asymptotic notations, you need to specify with regard to what point the asymptotics is taken. Here, it looks like this is when $x\to 0^+$; note that this could have equally been $x\to\infty$, so specifying it is required. Now, you have that for any fixed $\alpha > 0$, $$ \ln x = O(x^{-\alpha}) $$ when $x\to 0^+$, since ...


2

Use: 1) $\log_a X - \log_a Y=\log_a \frac{X}Y$ 2) $\log_a a=1$ $$\log_a (x^{a}-x)-\log_a \Big(\dfrac{x^{a}-x}{a}\Big)=$$ $$=\log_a\frac{(x^{a}-x)}{\Big(\dfrac{x^{a}-x}{a}\Big)}=\log_a a=1$$


2

Hint: There's a fundamental inequality involving integrals that says $$\left | \int_a^b f(t)\, dt \right | \leq (b-a) \cdot \max\{\left | f(t) \right |\,; \, t \in (a,b)\}$$ Can you think of how to use it?


2

it is $$\frac{\ln(x)}{\ln(2)}\frac{\ln(x)}{2\cdot\ln(2)}\frac{\ln(x)}{3\ln(2)}=\frac{9}{2}$$ and from here we get $$\frac{\ln(x)^3}{6\cdot (\ln(2))^3}=\frac{9}{2}$$ calculating this we obtain $$\ln(x)^3=27\cdot \ln(2)^3$$ can you proceed?


2

To systematically attack such questions, use $\log_x y = \frac{\log y}{ \log x}$ i.e. $$\log_3 8 = \frac{\ln 8}{\ln 3}, \quad \log_5 9 = \frac{\ln 9}{\ln 5}, \quad \log_2 5 = \frac{\ln 5}{\ln 2}$$ multiply and simplify $$\log_3 8 \times \log_5 9 \times \log_2 5 = \frac{\ln 8}{\ln 3} \times \frac{\ln 9}{\ln 5} \times \frac{\ln 5}{\ln 2}\\ =\frac{\ln 8 ...


2

Partial answer: As said, before extending the work, lets work out the ordinary case first. Let $X_1 = e^{\sigma Z_1}, X_2 = e^{\sigma Z_2}$ where $Z_1, Z_2$ are i.i.d. standard normal. So $$ \begin{align} &~ E[\min\{e^{\sigma Z_1}, e^{\sigma Z_2}\}] \\ =&~ E[e^{\sigma Z_1}\mathbf{1}_{Z_1 < Z_2}] + E[e^{\sigma Z_2}\mathbf{1}_{Z_2 < Z_1}]\\ ...


2

First of all you must know that whenever the base of a log is not mentioned then its base is 10. $$m\log a= \log(a^m)$$ It is wrong to assume it means "$\log 8$ to the base $2$". The correct value will be $2.4082…$ not $3$. You might get a value of 3 in that calculator because the base was not set to 10 or due to some other mistake.


2

I would choose to clarify with the person what does he mean. I could have interpreted it as $8\log_{10}2$ or $8 \ln2$ or $8\log_22$.


2

Using integration by parts, the original integral turns into: $$ \int_{0}^{1}\frac{x-1}{1-x+\frac{x^2}{2}}\,\log(x)\,dx \tag{1}$$ We may compute the Taylor series of $\frac{x-1}{1-x+\frac{x^2}{2}}$ in a neighbourhood of the origin through partial fraction decomposition, then exploit $$ \int_{0}^{1} x^k\log(x)\,dx = -\frac{1}{(k+1)^2}.\tag{2} $$ That ...


2

Your calculation is perfect. You're getting the answer as 56.893449... , this means that the value of car drops down to £5000 after 56th month not in the 56th month. This implies the answer is during 57th month. To understand this an analogy may be used. The year 2016 is in 21st century not 20th century.


1

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1

8log 2 means , find [log (base 10) 2] and then multiply the answer with 8.


1

I believe if some says "8 Log2" just means the logarithm of 8 * Log(2), otherwise if he intents to specify the base as 2 he should say "8's Log in base 2" that actually is 3.


1

Hint: Use the change of base formula to see $$ \log_4 x = {\log_2 x\over \log_2 4} = {\log_2 x \over 2} $$ Similarly $$ \log_8 x = {\log_2 x\over 3} $$


1

Note that if $1\le t\le x$, then $\frac{1}{x}\le\frac{1}{t}\le 1$. Therefore, we find that $$\frac{x-1}{x}\le \int_1^x \frac{1}{t}\,dt \le x-1 \tag 1$$ In fact, $(1)$ is true for all $x>0$. Finally, we have $$\frac{x-1}{x^3}\le \frac{1}{x^2}\int_1^x \frac{1}{t}\,dt\le \frac{x-1}{x^2}<\frac1x$$ for all $x>0$


1

Use: $$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$ When $a$ and $b$ are positive: $$\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$$ So, we get when $a$ and $x$ are positive ($x\ne1$ and $a\ne1$): $$\log_a\left(x^a-x\right)-\log_a\left(\frac{x^a-x}{a}\right)=\frac{\ln\left(x^a-x\right)}{\ln(a)}-\frac{\ln\left(x^a-x\right)-\ln\left(a\right)}{\ln(a)}=$$ ...


1

For $r>0$ take $M>1$ such that $1/M<r.$ For $y>M$ we have $$0<\ln y=\int_1^y (1/z)\;dz= \int_1^M(1/z)\;dz+\int_M^y(1/z)\;dz=$$ $$=\ln M+\int_M^y(1/z)\;dz<\ln M+\int_M^y(1/M)\;dz=\ln M+(y-M)/M.$$ $$\text { So, }\quad 0<(\ln y)/y<(\ln M)/y+(1-M/y)/M.$$ Therefore $$0\leq \sup_{x\geq y}\;(\ln x)/x\leq \sup_{x\geq y}(\;(\ln ...


1

$$ \ln\frac{(k+1)(k+2)}{k(k+3)}=\ln\frac{k^2+3k+2}{k^2+3k}=\ln\left(1+\frac2{k^2+3k}\right)\lt\frac2{k^2+3k}\;. $$ Thus \begin{align} \sum_{k=0}^\infty\left(-\ln(j+4k)+\ln(j+4k+1)+\ln(j+4k+2)-\ln(j+4k+3)\right) &\lt\sum_{k=0}^\infty\frac2{(j+4k)^2+3(j+4k)} \\ &\le\sum_{k=0}^\infty\frac2{(1+4k)^2+3(1+4k)} \\ &= \frac\pi{12}+\frac{\ln2}2 \\ ...



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