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7

The OP's start was fine. We rewrite the left side as $x+8$, and pull the exponent out of the right side, to get $$x+8=(x-1)\log_64$$ We expand the right side to get $$x+8=x(\log_64)-\log_64$$ then subtract $8+x\log_64$ from both sides, to get $$x-x\log_64=-8-\log_64$$ We now factor out $x$ to get $$x(1-\log_64)=-8-\log_64$$ Lastly, we divide by $1-\log_64$ ...


6

No, you haven't done anything wrong. $\ln 4$ is a constant. Hence its derivative is $0$.


5

$\sum_{k=1}^n\ln^2k\ne\ln^2n!$ because $$\ln^2(xy)=(\ln x+\ln y)^2\ne\ln^2x+\ln^2y.$$


3

So, let's say by "radical" you mean an element of the field $$\large F=\mathbb{Q}(\{a^{1/r}:a,r\in\mathbb{N}\}),$$ or if you want to allow "radicals" of negative integers, an element of the larger field $$\large L=F(\{e^{2\pi i/n}:n\in\mathbb{N}\}).$$ Then no, there are many logarithms with "radical" base and argument that are not themselves "radicals". ...


3

If you take $\log_a$, there are two possibilities. If $a>1$ then $\log_a$ is increasing, and the direction of the inequality does not change. If $0<a<1$, the function $\log_a$ is decreasing, and the direction of the inequality is reversed.


3

The question is: Which logarithm function are you using? If you are using the real-valued $\ln \colon (0,+\infty) \to \mathbb{R}$, your integrand is simply not defined on the interval $\bigl[0,\frac{\pi}{2}\bigr]$, and hence the integral neither. I put it into Alpha and it gave me -1/2 as answer. Unless you tell it not to, Wolfram Alpha happily uses a ...


3

It doesn't necessarily matter which base you are in, you can solve this as follows $$6^{x+8}=4^{x-1}\iff \log(6^{x+8})=\log(4^{x-1})$$ then, by rules of the logarithm, $$(x+8)\log 6=(x-1)\log 4$$ now, by the distributive property for real numbers, $$(x+8)\log 6=(x-1)\log 4 \iff x\log 6 + 8\log 6 = x\log 4 -\log4$$ and so $$x\log6-x\log 4=-8\log 6 - \log4$$ ...


2

I would ask that "someone" for more details first why this should be true. Forget about the logarithm; if its arguments are well-defined (positive), the inequality holds (assuming it should hold for any $h$) if and only if $\Sigma\preceq I$. But take, e.g., for $\rho_1=\rho_2=\rho_3=1/4$ and $h=[1,1]^T$, $$ h^*\Sigma ...


2

$$\log_x(y)+\log_y(x)=\frac{\ln(y)}{\ln(x)}+\frac{\ln(x)}{\ln(y)}=\frac{\ln^2(x)+\ln^2(y)}{\ln(x)\ln(y)}=\frac{(\ln(x)+\ln(y))^2}{\ln(x)\ln(y)}-2=\frac{\ln^2(xy)}{\ln(x)\ln(y)}-2$$ Then if $xy=64$, $$\log_x(y)+\log_y(x)=\frac{5}{2}\implies \frac{\ln^2(64)}{\ln(x)\ln(y)}=\frac{9}{2}\implies ...


2

If $$y=x^{1/2}$$ and $$xy=64$$ then $$x^{3/2}=64 = 2^6$$ so $$x=2^{6\times 2/3} = 2^4 =16$$ and $$y=16^{1/2}=4.$$


1

Hint: Plot the expression $|x+1|+|x-5|$. You will see that for $x \leq -1$, it is a line with slope $-2$; for $-1 \leq x \leq 5$, it is a horizontal line at $y = 6$; for $x \geq 5$, it is a line with slope $2$. It looks a bit like a playground swing. For any given $a$, the locations at which $\log_a x = |x+1|+|x-5|$ are the places where the graph of ...


1

For $0< x <5$, $$ \log_ax = x+1-x+5=6 $$ So, $$ a=\sqrt[6]{x} $$ Therefore, $$ 0<a<5^\dfrac{1}{6}\tag{1} $$ For $x>5$, $$ \log_ax = x+1+x-5=2x-4 $$ $$ \log_ax=2x-4 $$ $$ a=\sqrt[2x-4]{x} $$ Using graph, $$ 1<a<5^\dfrac{1}{6}\tag{2} $$ At $x=5$, $$ \log_a5 = 1+5=6 $$ Therefore, $$ a=5^{\frac 16} $$ From equation 1 and 2, $a=5^{\frac ...


1

This is just the change of base formula that has been rearranged. Let:$$\log_sa=x\tag{1}$$$$\therefore a=s^x$$Now takes logs of both sides to base $r$ to get:$$\log_ra=\log_r(s^x)=x\log_rs\tag{2}$$Now substitute (1) into (2) and you are done.


1

Presumably, this is your argument: $$2^{x-1}7^{\frac{1}{x}-1} \leq 1$$ $$2^{x-1}\leq 7^{1-\frac{1}{x}}$$ So: $$2^{x-1}\leq \left(7^{1/x}\right)^{x-1}\tag{1}$$ Now there should be two cases, one where $x-1<0$ and one where $x-1\geq 0$. If $x-1<0$, then $2\geq 7^{1/x}$, or $\log_7{2}\leq \frac{1}x$. Again there are two cases: a. If $x< 0$, you ...


1

If $x<1$ the function $f(t)=t^{1/(x-1)}$ is decreasing for $t>0$. This means that if you raise both sides of inequality to $\frac1{x-1}$, its direction is reversed. If I have understood you correctly, you have arrived at $$2^{x-1}\le 7^{(x-1)/x}$$ Now, you should take logarithms in base $2$ (which is an increasing function) to get ...


1

Consider the function $$f(x)=x\log(x)+px - q$$ The first derivative $$f'(x)=\log (x)+1+p$$ cancels at a single place $x=e^{-(p+1)}$ and at this point the value of the function is $-(q+e^{-(p+1)})$. The second derivative being always positive, then this point corresponds to a minimum. So, the function (which is defined for $x \geq 0$) starts from $-q$, goes ...


1

Remember the rule for derivatives of $\ln$: $$\frac {d}{dx} \ln(f(x)) = \frac {\frac {d}{dx}f(x)}{f(x)}$$ In your case $f(x)=4$, thus $$\frac {d}{dx} \ln(4) = \frac {\frac {d}{dx} 4}{4} = \frac {0}{4} = 0$$


1

One should be careful about the exponent, there are two possible meanings: \begin{equation*} \ln \left( \left( \frac{1}{x}\right) ^{x}\right) \neq \left( \ln \left( \frac{1}{x}\right) \right) ^{x} \end{equation*} Let us compute the limit of both of them. \begin{equation*} \lim_{x\rightarrow 0^{+}}\ln \left( \left( \frac{1}{x}\right) ^{x}\right) ...


1

$$\lim_{x\to 0}\ln\left(\frac{1}{x}\right)^x\neq \lim_{x\to 0}x\ln\left(\frac{1}{x}\right)$$ You have that $$\ln\left(\frac{1}{x}\right)^x=e^{x\ln(\frac{1}{x})}.$$ Since $$\lim_{x\to 0}x\ln\left(\frac{1}{x}\right)=\lim_{x\to 0}\frac{\ln\left(\frac{1}{x}\right)}{\frac{1}{x}}\underset{\text{(Hop.)}}{=}\lim_{x\to ...


1

This doesn't make much sense, if the interest is evaluated only once a year. Clearly, after 14 years you would only have $1.05^{14}=1.9799\ldots$ times the initial amount and only after 15 years you will have twice the initial amount (and a little bit more). I think there is an error in the book, unless it is understood that the interest is evaluated ...



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