How do I solve logarithms with addition in them?

$2.5({e}^{c\cdot 100}+{e}^{-c\cdot 100})=30$

$2.5({e}^{c\cdot 100}+{e}^{-c\cdot 100})=30$

Diego Barr
2022-10-22
Answered

How do I solve logarithms with addition in them?

$2.5({e}^{c\cdot 100}+{e}^{-c\cdot 100})=30$

$2.5({e}^{c\cdot 100}+{e}^{-c\cdot 100})=30$

You can still ask an expert for help

Milton Hampton

Answered 2022-10-23
Author has **16** answers

let

$$a={e}^{100c}$$

then we have

$$a+\frac{1}{a}=12$$

and you will get a quadratic equation in $a$.

$$a={e}^{100c}$$

then we have

$$a+\frac{1}{a}=12$$

and you will get a quadratic equation in $a$.

asked 2022-09-24

Upperbound for $\sum _{i=1}^{N}{a}_{i}\mathrm{ln}{a}_{i}$

It's easy to prove that following upperbound is true:

$\sum _{i=1}^{N}{a}_{i}\mathrm{ln}{a}_{i}\le A\mathrm{ln}A$, where $\sum _{i=1}^{N}{a}_{i}=A$ and ${a}_{i}\ge 1$

I'm wondering, is there stronger upperbound?

It's easy to prove that following upperbound is true:

$\sum _{i=1}^{N}{a}_{i}\mathrm{ln}{a}_{i}\le A\mathrm{ln}A$, where $\sum _{i=1}^{N}{a}_{i}=A$ and ${a}_{i}\ge 1$

I'm wondering, is there stronger upperbound?

asked 2022-06-26

How is natural log integration broken up into this range? (equation is contained the script)

When I was reading a paper, I found an strange derivation like

${\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}\mathrm{l}\mathrm{n}(1+{e}^{w})f(w)dw\phantom{\rule{0ex}{0ex}}={\int}_{-\mathrm{\infty}}^{0}\mathrm{ln}(1+{e}^{w})f(w)+{\int}_{0}^{\mathrm{\infty}}[\mathrm{ln}(1+{e}^{-w})+w]f(w)dw$

when $w$ is the normal random variable and $f(w)$ is the normal density.

Why is that natural log integration broken up into that?

I think it just need a simple principle.

Thank you.

When I was reading a paper, I found an strange derivation like

${\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}\mathrm{l}\mathrm{n}(1+{e}^{w})f(w)dw\phantom{\rule{0ex}{0ex}}={\int}_{-\mathrm{\infty}}^{0}\mathrm{ln}(1+{e}^{w})f(w)+{\int}_{0}^{\mathrm{\infty}}[\mathrm{ln}(1+{e}^{-w})+w]f(w)dw$

when $w$ is the normal random variable and $f(w)$ is the normal density.

Why is that natural log integration broken up into that?

I think it just need a simple principle.

Thank you.

asked 2022-04-07

Find the solution for the exponential equation in terms of logarithms. $\frac{8x}{2}={5}^{1-x}$

asked 2021-10-13

Solve the exponential equation $5}^{2x+3}={3}^{x-1$ . Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution.

asked 2022-11-20

Numbers of solutions of the equation ${\mathrm{log}}_{3}\frac{2{x}^{2}+3x+3}{5}=\frac{1}{{\mathrm{log}}_{2{x}^{2}+3x+9}9}$

Pretty straightforward question. When I solved it, I got two positive and two negative solutions, so that would make 4 in total. None get discarded as the arguments in the logarithm still stay positive.

However, the solution is supposed to be 2, so I'm not sure where I went wrong.

I got the values of $x$ as $-2$, $1/2$, $12/5$ and $-18/5$

Pretty straightforward question. When I solved it, I got two positive and two negative solutions, so that would make 4 in total. None get discarded as the arguments in the logarithm still stay positive.

However, the solution is supposed to be 2, so I'm not sure where I went wrong.

I got the values of $x$ as $-2$, $1/2$, $12/5$ and $-18/5$

asked 2022-08-15

Exponential continuous growth $\mathrm{ln}a$ vs. $r$? Huh?

So, given a simple population continuous growth problem, it seems that the entirety of the internet uses $P={P}_{0}{e}^{rt}$ where $P$ is the population over time, ${P}_{0}$ is the initial population, $r$ is the percentage of change, and $t$ is time.

But this class is asserting that $P={P}_{0}{e}^{(\mathrm{ln}a)(t)}$ where $a=1+r$, and everything else stays the same.

So lets say a population starts at $10,000$ with $10$ continuous growth rate. What is the population in $10$ years?

The internet says that $P=10,000{e}^{(.10)(10)}=27,182$

But the class says that $P=10,000{e}^{(\mathrm{ln}1.10)10}=25,937$ ($a=1+r$ which is $1.10$)

but that equals the usual constant growth of $P={P}_{0}{a}^{t}=25937$

What the heck? What am I missing?

So, given a simple population continuous growth problem, it seems that the entirety of the internet uses $P={P}_{0}{e}^{rt}$ where $P$ is the population over time, ${P}_{0}$ is the initial population, $r$ is the percentage of change, and $t$ is time.

But this class is asserting that $P={P}_{0}{e}^{(\mathrm{ln}a)(t)}$ where $a=1+r$, and everything else stays the same.

So lets say a population starts at $10,000$ with $10$ continuous growth rate. What is the population in $10$ years?

The internet says that $P=10,000{e}^{(.10)(10)}=27,182$

But the class says that $P=10,000{e}^{(\mathrm{ln}1.10)10}=25,937$ ($a=1+r$ which is $1.10$)

but that equals the usual constant growth of $P={P}_{0}{a}^{t}=25937$

What the heck? What am I missing?

asked 2022-11-21

Equations transformations with roots

How does the following transformation works (do not write that it is easy i want the answer):

$\mathrm{ln}\sqrt[n]{\frac{n!}{{n}^{n}}}=\frac{\mathrm{ln}\frac{n!}{{n}^{n}}}{n}$

How does the following transformation works (do not write that it is easy i want the answer):

$\mathrm{ln}\sqrt[n]{\frac{n!}{{n}^{n}}}=\frac{\mathrm{ln}\frac{n!}{{n}^{n}}}{n}$