Show that $(\mathrm{ln}a{)}^{k}\ne k\mathrm{ln}a$

Widersinnby7
2022-11-21
Answered

Show that $(\mathrm{ln}a{)}^{k}\ne k\mathrm{ln}a$

You can still ask an expert for help

meexeniexia17h

Answered 2022-11-22
Author has **18** answers

Because we need to disprove this statement, we only need to find 1 counterexample.

$(\mathrm{log}e{)}^{2}\ne 2(\mathrm{log}e)$

Statement proved.

$(\mathrm{log}e{)}^{2}\ne 2(\mathrm{log}e)$

Statement proved.

asked 2022-09-23

Product of logarithms, prove this identity.

Is it hard to prove this identity:

$2\mathrm{log}(a)\mathrm{log}(b)=\mathrm{log}(ab{)}^{2}-\mathrm{log}(a{)}^{2}-\mathrm{log}(b{)}^{2}$

for $a>1$ and $b>1$?

Is it hard to prove this identity:

$2\mathrm{log}(a)\mathrm{log}(b)=\mathrm{log}(ab{)}^{2}-\mathrm{log}(a{)}^{2}-\mathrm{log}(b{)}^{2}$

for $a>1$ and $b>1$?

asked 2022-11-14

Logarithmic Differentiation equation, Help!

So, I have to differentiate this via $\mathrm{log}$. I am still learning, so please be patient, I will try to explain everything I did. Please tell me if it is correct.

$y=\frac{(x+3{)}^{4}(2{x}^{2}+5x{)}^{3}}{\sqrt{4x-3}}$

$\mathrm{ln}(y)=\mathrm{ln}((x+3{)}^{4}(2{x}^{2}+5x{)}^{3}))-\mathrm{ln}\left(\sqrt{4x-3}\right)$

$\mathrm{ln}(y)=4\mathrm{ln}(x+3)3\mathrm{ln}(2{x}^{2}+5x)-\frac{1}{2}\mathrm{ln}(4x-3)$

aaaaaaaaand don't know what to do next, any help in the process or next step?

So, I have to differentiate this via $\mathrm{log}$. I am still learning, so please be patient, I will try to explain everything I did. Please tell me if it is correct.

$y=\frac{(x+3{)}^{4}(2{x}^{2}+5x{)}^{3}}{\sqrt{4x-3}}$

$\mathrm{ln}(y)=\mathrm{ln}((x+3{)}^{4}(2{x}^{2}+5x{)}^{3}))-\mathrm{ln}\left(\sqrt{4x-3}\right)$

$\mathrm{ln}(y)=4\mathrm{ln}(x+3)3\mathrm{ln}(2{x}^{2}+5x)-\frac{1}{2}\mathrm{ln}(4x-3)$

aaaaaaaaand don't know what to do next, any help in the process or next step?

asked 2022-11-04

Solving logarithmic equation $2\mathrm{log}(x)+1=\mathrm{log}(19x+2)$

I know the solution has to be $x=2$

However I can't find the manual steps (Wolfram doesn't know the manual steps either).

This is all I got

$$\mathrm{log}({x}^{2})+1=\mathrm{log}(19x+2)$$

$$\mathrm{log}({x}^{2})-\mathrm{log}(19x+2)=-1$$

$$\mathrm{log}\left(\frac{{x}^{2}}{19x+2}\right)=-1$$

I know the solution has to be $x=2$

However I can't find the manual steps (Wolfram doesn't know the manual steps either).

This is all I got

$$\mathrm{log}({x}^{2})+1=\mathrm{log}(19x+2)$$

$$\mathrm{log}({x}^{2})-\mathrm{log}(19x+2)=-1$$

$$\mathrm{log}\left(\frac{{x}^{2}}{19x+2}\right)=-1$$

asked 2022-07-31

$\mathrm{ln}(1/25)$ in terms of $\mathrm{ln}5\text{}and\text{}\mathrm{ln}7$

asked 2022-11-12

Is there a simple algorithm for exponentiating large numbers to large powers?

I've been thinking about this for some days, a multiplication is a lot of sums, so:

$75\times 75=\stackrel{\text{75 times}}{\stackrel{\u23de}{75+75+75+75+75+75+75+75+\cdots}}$

But then, there is a simple algorithm that enable us to multiply without having to sum all those numbers. In the same way, exponentiation is repeated multiplication - but I wasn't taugth about such algorithm for exponentiation. I've been thinking in representing the number with a polynomial, for example:

$1038={10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}$

Then:

${1038}^{1038}=({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}{)}^{1038}$

But from here (considering what I currently know) I'd have to multiply it $1038$ times. The mentioned multiplication would be:

$({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}{)}^{1038}=\stackrel{\text{1038 times}}{\stackrel{\u23de}{({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0})\cdot ({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0})\cdot \dots}}$

The first idea I had: There might be some connection with the binomial theorem, but I don't see how it fits. The second idea would be to find a way to write:

$({10}^{{3}}+3\cdot {10}^{{1}}+8\cdot {10}^{{0}}{)}^{1038}$

In which the red exponents are multiplied in some way by $1038$. There might be some connection with logarithms here, but I don't see it. And it could be the case that these techniques won't yield the results I'm looking for, so: Is there a simple algorithm for large numbers elevated to large exponents?

I've been thinking about this for some days, a multiplication is a lot of sums, so:

$75\times 75=\stackrel{\text{75 times}}{\stackrel{\u23de}{75+75+75+75+75+75+75+75+\cdots}}$

But then, there is a simple algorithm that enable us to multiply without having to sum all those numbers. In the same way, exponentiation is repeated multiplication - but I wasn't taugth about such algorithm for exponentiation. I've been thinking in representing the number with a polynomial, for example:

$1038={10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}$

Then:

${1038}^{1038}=({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}{)}^{1038}$

But from here (considering what I currently know) I'd have to multiply it $1038$ times. The mentioned multiplication would be:

$({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}{)}^{1038}=\stackrel{\text{1038 times}}{\stackrel{\u23de}{({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0})\cdot ({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0})\cdot \dots}}$

The first idea I had: There might be some connection with the binomial theorem, but I don't see how it fits. The second idea would be to find a way to write:

$({10}^{{3}}+3\cdot {10}^{{1}}+8\cdot {10}^{{0}}{)}^{1038}$

In which the red exponents are multiplied in some way by $1038$. There might be some connection with logarithms here, but I don't see it. And it could be the case that these techniques won't yield the results I'm looking for, so: Is there a simple algorithm for large numbers elevated to large exponents?

asked 2021-11-08

Rewrite the following logarithms in expanded form by applying the properties of logarithms.

$\mathrm{log}\left({x}^{y}\right)$

asked 2022-10-16

Calculate $\mathrm{ln}97$ and ${\mathrm{log}}_{10}97$ without calculator accurate up to 2 decimal places.

I have rote some value of logs of prime numbers up to $11$

$97$ is a little big.

In case it would have been a multiple of smaller primes, I would have used the trick of logarithm identities .

But I am confused how to do it, or will it be ok to approximate it to 96 or 98.

I also don't know much calculus except differentiation and integration.

I look for a short and simple way.

I have studied maths up to 12th grade.

I have rote some value of logs of prime numbers up to $11$

$97$ is a little big.

In case it would have been a multiple of smaller primes, I would have used the trick of logarithm identities .

But I am confused how to do it, or will it be ok to approximate it to 96 or 98.

I also don't know much calculus except differentiation and integration.

I look for a short and simple way.

I have studied maths up to 12th grade.