I just wonder can we call ${2}^{x}+x$ an exponential function or not?

With the same way of thinking is ${\mathrm{log}}_{2}(x)+x$ a logarithmic?

With the same way of thinking is ${\mathrm{log}}_{2}(x)+x$ a logarithmic?

Braylon Lester
2022-07-22
Answered

I just wonder can we call ${2}^{x}+x$ an exponential function or not?

With the same way of thinking is ${\mathrm{log}}_{2}(x)+x$ a logarithmic?

With the same way of thinking is ${\mathrm{log}}_{2}(x)+x$ a logarithmic?

You can still ask an expert for help

asked 2022-07-20

$y=(3{x}^{2}+2{)}^{lnx}$

The answer to this is:

$\frac{dy}{dx}=(3{x}^{2}+2{)}^{lnx}(\frac{1}{x}ln(3{x}^{2}+2)+\frac{6xlnx}{3{x}^{2}+2})$

What I'm coming up with is:

$\frac{dy}{dx}=(3{x}^{2}+2{)}^{lnx}(\frac{1}{x}ln(3{x}^{2}+2)+\frac{6x}{3{x}^{2}+2})$

What I'm not understanding is where the $\frac{6xlnx}{3{x}^{2}+2}$ comes from, if anyone could explain this I'd really appreciate it.

The answer to this is:

$\frac{dy}{dx}=(3{x}^{2}+2{)}^{lnx}(\frac{1}{x}ln(3{x}^{2}+2)+\frac{6xlnx}{3{x}^{2}+2})$

What I'm coming up with is:

$\frac{dy}{dx}=(3{x}^{2}+2{)}^{lnx}(\frac{1}{x}ln(3{x}^{2}+2)+\frac{6x}{3{x}^{2}+2})$

What I'm not understanding is where the $\frac{6xlnx}{3{x}^{2}+2}$ comes from, if anyone could explain this I'd really appreciate it.

asked 2022-07-18

A point on a graph is (1/8,−3) of the logarithmic function $f(x)=\mathrm{log}{b}^{x}$, and the point (4,k) is on the graph of the inverse, $y={f}^{-1}(x)$. Determine the value k.

asked 2022-07-21

Help me with this

$\int \frac{{\mathrm{ln}}^{3}x}{x}\text{}dx$

Is this the same as

$\int \frac{(\mathrm{ln}x{)}^{3}}{x}\text{}dx\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}?$

$\int \frac{{\mathrm{ln}}^{3}x}{x}\text{}dx$

Is this the same as

$\int \frac{(\mathrm{ln}x{)}^{3}}{x}\text{}dx\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}?$

asked 2022-08-11

We know that:

${e}^{x}=\sum _{n=0}^{\mathrm{\infty}}\frac{{x}^{n}}{n!}$

$\mathrm{log}(x)=\sum _{n=1}^{\mathrm{\infty}}\frac{(-1{)}^{n+1}(x-1{)}^{n}}{n}$

Where second series converges when |x−1|<1.It is possible to prove that:

${e}^{\mathrm{log}(x)}=x$

for $|x-1|<1$ using only series representation?

${e}^{x}=\sum _{n=0}^{\mathrm{\infty}}\frac{{x}^{n}}{n!}$

$\mathrm{log}(x)=\sum _{n=1}^{\mathrm{\infty}}\frac{(-1{)}^{n+1}(x-1{)}^{n}}{n}$

Where second series converges when |x−1|<1.It is possible to prove that:

${e}^{\mathrm{log}(x)}=x$

for $|x-1|<1$ using only series representation?

asked 2022-09-23

Given a constant sum of ${x}_{n}$ values:

$\sum _{n=1}^{N}{x}_{n}=C$

where ${x}_{n}\ge 1$

Find the maximum of the following expression:

$\sum _{n=1}^{N}{a}_{n}\mathrm{log}\left({x}_{n}\right)$

where the ${a}_{n}>0$ values are constants, and the ${x}_{n}$ values are free to change, given that their sum will remain constant.

$\sum _{n=1}^{N}{x}_{n}=C$

where ${x}_{n}\ge 1$

Find the maximum of the following expression:

$\sum _{n=1}^{N}{a}_{n}\mathrm{log}\left({x}_{n}\right)$

where the ${a}_{n}>0$ values are constants, and the ${x}_{n}$ values are free to change, given that their sum will remain constant.

asked 2022-08-06

The sum of series $\frac{(\mathrm{log}3{)}^{1}}{1!}+\frac{(\mathrm{log}3{)}^{3}}{3!}+\frac{(\mathrm{log}3{)}^{5}}{5!}+\cdots $ is what? Is there a general algorithm to find the summation of logarithms?

asked 2022-08-10

I'm confused trying to figure out the domain of the logarithmic function below:

$f(x)=\mathrm{ln}({e}^{x}+3)$

Because the argument of $f,\phantom{\rule{thinmathspace}{0ex}}{e}^{x}+3,$, is a nonnegative number, ${e}^{x}+3>0$ and ${e}^{x}>-3.$ Taking natural logarithms on both sides, we get $x>\mathrm{ln}(-3)$. However, the domain of a logarithmic function must be nonnegative real numbers, so $\mathrm{ln}(-3)$ doesn't make sense. How then to determine the scope of the original function?

$f(x)=\mathrm{ln}({e}^{x}+3)$

Because the argument of $f,\phantom{\rule{thinmathspace}{0ex}}{e}^{x}+3,$, is a nonnegative number, ${e}^{x}+3>0$ and ${e}^{x}>-3.$ Taking natural logarithms on both sides, we get $x>\mathrm{ln}(-3)$. However, the domain of a logarithmic function must be nonnegative real numbers, so $\mathrm{ln}(-3)$ doesn't make sense. How then to determine the scope of the original function?