6

Let's start with your second definition. We want to find the derivative of $f(x) = a^x$, for some $a>0.$ Using the definition of derivative, we get \begin{align} f'(x) &= \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}\\ &= \lim_{h\to 0}\frac{a^{x+h} - a^{x}}{h}\\ &=a^{x}\lim_{h\to 0}\frac{a^{h} - 1}{h}. \end{align} Now, as you said, one way to define $...


5

This Wikipedia article: Characterizations of the exponential function, shows six ways to define the exponential function $e^x$ (thus for the constant $e$) and a detailed discussion of equivalency of all the six definitions. This happens a lot in mathematics. The same mathematical object may have many properties that can characterize the object. Once they are ...


4

$\ln (xy)=\ln (x)+\ln (y)$ holds only for positive numbers $x$ and $y$. You cannot use negative numbers in this.


3

The absolute value is present exactly to avoid this. When you write $f(x) = ln|x|$ It means that a negative number cannot come inside the bracket. so when you apply the second form of function, x is a negative number and negative and negative cancel out. What you have done is similar to $ln(1)=ln(-1*-1)=ln(-1)+ln(-1)$, which cannot be done, as the property ...


3

Let $z$ be your right-hand side. Then some manipulations (divide by $a$, exponentiate, divide by $a$ again) yields $$x+a \log x=z\implies (x/a)e^{x/a} = e^{z/a}/a\implies x/a = W(e^{z/a}/a)$$ where $W$ is the Lambert-W function. So $x=a W(e^{z/a}/a)$.


3

Note that\begin{align*}\ln(x^2-10)+\ln(9)=\ln(10)&\iff\ln\bigl(9(x^2-10)\bigr)=\ln(10)\\&\iff9x^2-90=10\\&\iff x^2=\frac{100}9\\&\iff x=\pm\frac{10}3.\end{align*} The error from your first attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln(10)\iff x^2-10+9=10,$$whereas the error in your second attempt lies in assuming that$$\ln(x^2-10)+\ln(...


2

By some easy algebraic computations, we get $x^{\log_2x}=2^{\log_2(x^{\log_2x})}=2^{\log_2x \times \log_2x}=2^{\log_2^2x}$. $\Rightarrow \log_2^2x = 4$ $\Rightarrow \log_2x = \pm 2$ $\Rightarrow x_1 = 2^2 = 4$ and $x_2 = 2^{-2} = \frac{1}{4}$


2

Your first attempt simply uses the strategy "let's ignore $\ln$": but it does not work. The second attempt is better, but you made a mistake. Actually $$\ln (x^2-10) + \ln 9$$ becomes $$\ln ((x^2-10) \cdot 9)$$ then you have $$(x^2-10) \cdot 9=10$$ which is easily solved.


2

There is a formula to change the base of logarithms $$\log_a b=\frac{\log_c b}{\log_c a}$$ for instance $$\log_{20} 100+\log_{100} 20=\frac{\log_{100} 100}{\log_{100} 20}+\log_{100} 20=\frac{1}{\log_{100} 20}+\log_{100} 20$$ Note that $$\log_{100}x=\frac{\log_{10}x}{\log_{10}100}=\frac{1}{2}\log_{10}x$$


2

For the reciprocal of the logarithm of $x$ to be defined, the logarithm of $x$ must be defined in the first place. That means that we need to require $x>0$: The logarithm s neither defined for $0$ nor for negative numbers. In addition, taking the reciprocal must be defined, which is only defined if we do not attempt to take the reciprocal of $0$. As $\log ...


2

As @Semiclassical wrote, $$x+a\log(x)=k \implies x=a W\left(\frac{1}{a}e^{\frac{k}{a}}\right)$$ If you cannot use Lambert function, you will need a numerical method and take into account the fact that $\forall t \geq e$ $$W(t) \geq\log (t)-\log (\log (t))+\frac 12\frac{\log (\log (t))}{ \log (t)}=f(t)$$ $$W(t) \leq \log (t)-\log (\log (t))+\frac e{e-1}\frac{\...


2

Hint: Between any two zeros there is a point where the derivative is $0$. But the derivative of $2x-\ln (1+x)$ has only one zero.


2

The equality $\ln (xy) = \ln x + \ln y$ only holds if all components are well defined, namely if $x,y >0$. The same happens in situations that do not involve the absolute value... For instance $\ln (x^2)$ is well defined for any $x \ne 0$, but the property $\ln(x^2) = 2 \ln x$ only holds for $x>0$.


2

Since the absolute value function isn't everywhere-differentiable, we will get rid of them. Luckily everything is continuous, so we can plug in $x=0$ to find out the signs we need. In particular, note that $\log(2)-\sin(0)$ is positive, and so $f(x)=\log(2)-\sin(x)$ (without the absolute value) in a neighborhood of zero. Hence, $g(x)=|\log(2)-\sin(\log(2)-\...


2

Hint: $\log a^b = b \log a$ and not $\log_ba$


1

One of the most important properties of logarithms is that $$ \log a+ \log b = \log ab \, . $$ In fact, if we try to solve the functional equation $$ f(a)+f(b) = f(ab) \tag{*}\label{*} \, , $$ then the solution is $f(x)=c\log x$, where $c$ is an arbitrary constant and $\log$ is the natural logarithm. Now consider the task of finding the area under the curve ...


1

I fact the integral can be evaluated by means of contour integrating in the complex plane. Let' consider the following closed contour C in the upper half-plane with the cut from $x=ia$ to $x=0$. We also choose the integrand $f(x)=\frac{\log(x^2+a^2)}{1+x^2}$ on the positive part of axis (right side). For the convenience of calculations we take $a<1$. $x=i$...


1

$$I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx= \int_0^{\infty} \frac{\ln(x^2+a^2)}{1+x^2}dx$$ $$\frac{dI}{da}=\int_0^{\infty} \frac{2a}{(1+x^2)(x^2+a^2)}dx$$ $$\int \frac{2a}{(1+x^2)(x^2+a^2)}dx=\frac{2}{a^2-1}\tan^{-1}(\frac{x}{a})-\frac{2a}{a^2-1}\tan^{-1}(x)$$ $$\frac{dI}{da}=\frac{2}{a^2-1}\frac{\pi}{2}-\frac{2a}{a^2-1}\frac{\pi}{2}$$ ...


1

Assuming we are talking about pointwise limits, we have that $$ \lim \frac 1n \ln(f_n(x,y)) = \lim \ln (f_n(x,y)^{1/n}) = \ln \left( \lim (f_n(x,y)^{1/n})\right) $$ So, what you can say is $$ \lim f_n(x,y)^{1/n} = e^{g(x,y)}. $$ Note: the interchange between $\ln$ and $\lim$ is justified by the continuity of the logarithm function.


1

Remember that in your exercise $\log$ means $\log_{10}$. So you have $\log(10)=1$. Now, \begin{align} \frac{3}{\log_2(10)}-\log(x-9)&=\log(44)\\ \log(x-9)&=\frac{3}{\log_2(10)}-\log(44)&\textrm{(put the unknown to one side)}\\ \log(x-9)&=\frac{3\log(2)}{\log(10)}-\log(44)&(\log_2(10)=\frac{\log(10)}{\log(2)})\\ \log(x-9)&=3\log(2)-\...


1

If $\log$ means $\log_{10}$ then your mistake is in this step $$\frac{3}{(\frac{\log(10)}{\log(2)})} = \frac{3}{(\log(10/2)}$$ You should have $\log_{10}-\log{2}$. By properties of $\log$ $$\log \left( \frac{a}{b}\right) = \log a - \log b$$


1

For $a,b>0$ we have $$a=b^x \iff \ln a = x \ln b.$$ can you proceed ?


1

Here is an alternative approach: if we first define the natural logarithm as $$ \log x = \int_{1}^{x}\frac{1}{t} \, dt $$ and the exponential function as the inverse of $\log$, then it is not much work to show that $e^x$ is its own derivative. (I go into more detail about this here.) Then, the limit $$ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n $$ can ...


1

Take the second definition: Consider: $\lim_{h\to 0}\frac {b^h -1}h = \ln b = 1$. And solve for $b$. That should be $e$ and we could define, supposedly, $e$ as the $b$ where that limit is $1$. .... $\lim_{h\to 0}\frac {b^h -1}h = $ $\lim_{\frac 1n\to 0}\frac {b^{\frac 1n} -1}{\frac 1n} = $ $\lim_{n\to \infty} n(b^{\frac 1n} - 1)=1$ Now for any $k$ if $k(...


1

As others have mentioned, there's no closed-form solution to $$x^{1/3} = (\log_2 x)^3$$ in elementary functions, but it can be solved using the Lambert W function. Briefly, the Lambert W function is the (multi-branched) inverse function of $$f(x) = xe^x$$ That is, $$x = W_k(x)e^{W_k(x)}$$ When working with real numbers, we only need the $W_{-1}$ and $W_0$ ...


1

Cubing, consider $$f(x)=x-\frac{\log ^9(x)}{\log ^9(2)}$$ $$f'(x)=1-\frac{9 \log ^8(x)}{x \log ^9(2)}$$ $$f''(x)=\frac{9 (\log (x)-8) \log ^7(x)}{x^2 \log ^9(2)}$$ The first derivative cancels at two points $$x_{1,2}=\exp\Big[-8 W(\pm k) \Big]\qquad \qquad k=\frac{\log ^{\frac{9}{8}}(2)}{8 \sqrt[4]{3}}$$ where $W(.)$ is Lambert function (numerically, $x_1=0....


1

Have you tried this? If not, take a look. Yes, there is no method to exactly compare them, unless you consider this special function $W_z(k)$, as a known function. https://www.wolframalpha.com/input/?i=x%5E%281%2F3%29-log2%28x%29%5E3 If you refer to the behavior of log/power to infinity (i.e. for x sufficiently large), always the power $x^n, n>0$ goes ...


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