Evaluate the expression without using a calculator.

$${\mathrm{log}}_{5}1$$

$${\mathrm{log}}_{5}1$$

Eliza Gregory
2022-10-17
Answered

Evaluate the expression without using a calculator.

$${\mathrm{log}}_{5}1$$

$${\mathrm{log}}_{5}1$$

You can still ask an expert for help

wlanauee

Answered 2022-10-18
Author has **17** answers

let us assume

$${\mathrm{log}}_{5}1=x$$

we know that $${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{b}x}{{\mathrm{log}}_{b}a}$$

so, $${\mathrm{log}}_{5}1=\frac{{\mathrm{log}}_{a}1}{{\mathrm{log}}_{a}5}=x\phantom{\rule{0ex}{0ex}}{\mathrm{log}}_{a}=x\cdot {\mathrm{log}}_{a}5\phantom{\rule{0ex}{0ex}}0=x{\mathrm{log}}_{a}5$$\

$${\mathrm{log}}_{5}1=x$$

we know that $${\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{b}x}{{\mathrm{log}}_{b}a}$$

so, $${\mathrm{log}}_{5}1=\frac{{\mathrm{log}}_{a}1}{{\mathrm{log}}_{a}5}=x\phantom{\rule{0ex}{0ex}}{\mathrm{log}}_{a}=x\cdot {\mathrm{log}}_{a}5\phantom{\rule{0ex}{0ex}}0=x{\mathrm{log}}_{a}5$$\

asked 2022-03-21

Log of negative numbers

I know that log of negative numbers is complex numbers. But I just got over this little proof and wondering what is wrong with this?

$\mathrm{log}(-a)=\frac{2\times \mathrm{log}(-a)}{2}=\frac{1}{2}\mathrm{log}\left({a}^{2}\right)=\frac{2}{2}\mathrm{log}\left(a\right)=\mathrm{log}\left(a\right)$

I know that log of negative numbers is complex numbers. But I just got over this little proof and wondering what is wrong with this?

asked 2022-08-31

How do I calculate the inverse function of this function?

I have this function:

$f(x)=\frac{1+\mathrm{ln}(x)}{1-\mathrm{ln}(x)}$

And i should calculate ${f}^{-1}(x)$

I am not really sure how to proceed but I think that the first step would be to have x alone, how do I achieve that?

I have this function:

$f(x)=\frac{1+\mathrm{ln}(x)}{1-\mathrm{ln}(x)}$

And i should calculate ${f}^{-1}(x)$

I am not really sure how to proceed but I think that the first step would be to have x alone, how do I achieve that?

asked 2022-11-10

A question in interview for trinity college, Cambridge

Let $M$ be a large real number. Explain why there must be exactly one root $w$ of the equation $Mx={e}^{x}$ with $w>1$. Why is log $M$ a reasonable approximation to $w$? Write $w=\mathrm{log}M+y$. Can you give an approximation to y, and hence improve on log $M$ as an approximation to $w$?

I think for the first part perhaps a graph suffices. But I feel a bit unsure of the meaning of $\mathrm{log}M$, is it $\mathrm{ln}M$? or ${\mathrm{log}}_{10}M$?

Thanks for your help in advance.

Let $M$ be a large real number. Explain why there must be exactly one root $w$ of the equation $Mx={e}^{x}$ with $w>1$. Why is log $M$ a reasonable approximation to $w$? Write $w=\mathrm{log}M+y$. Can you give an approximation to y, and hence improve on log $M$ as an approximation to $w$?

I think for the first part perhaps a graph suffices. But I feel a bit unsure of the meaning of $\mathrm{log}M$, is it $\mathrm{ln}M$? or ${\mathrm{log}}_{10}M$?

Thanks for your help in advance.

asked 2022-07-01

How to solve $q=\frac{\mathrm{ln}n}{\mathrm{ln}b+\mathrm{ln}q+\mathrm{ln}\mathrm{ln}n}$

asked 2022-07-03

Proof ${e}^{x}=\mathrm{exp}(x)$?

Define

$\mathrm{ln}(x)={\int}_{1}^{x}\frac{1}{t}$

Assume I have proven that $\mathrm{ln}x$ is one-to-one and therefore has an inverse $\mathrm{exp}(x)$

Define $e$ as:

$\mathrm{ln}e=1$

Now, if you have no other notion of exponentials, or logarithms, how could define what ${e}^{x}$ means and show that its the inverse of $\mathrm{ln}x$?

You are allowed to assume the logarithmic product and quotient property.

Thanks for the help.

Define

$\mathrm{ln}(x)={\int}_{1}^{x}\frac{1}{t}$

Assume I have proven that $\mathrm{ln}x$ is one-to-one and therefore has an inverse $\mathrm{exp}(x)$

Define $e$ as:

$\mathrm{ln}e=1$

Now, if you have no other notion of exponentials, or logarithms, how could define what ${e}^{x}$ means and show that its the inverse of $\mathrm{ln}x$?

You are allowed to assume the logarithmic product and quotient property.

Thanks for the help.

asked 2022-07-03

Showing $\frac{x}{1+x}<\mathrm{log}(1+x)<x$ for all $x>0$ using the mean value theorem

I want to show that

$\frac{x}{1+x}<\mathrm{log}(1+x)<x$

for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.

$\frac{x}{1+x}<\mathrm{log}(1+x)\iff \frac{x}{1+x}-\mathrm{log}(1+x)<0$

Let

$f(x)=\frac{x}{1+x}-\mathrm{log}(1+x).$

Since

$f(0)=0$

and

${f}^{\prime}(x)=\frac{1}{(1+x{)}^{2}}-\frac{1}{1+x}<0$

for all $x>0$, $f(x)<0$ for all $x>0$. Is this correct so far?

I go on with the second part: Let

$f(x)=\mathrm{log}(x+1)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an ${x}_{0}$ between $a$ and $x$ with

${f}^{\prime}({x}_{0})=\frac{f(x)-f(a)}{x-a}\iff \frac{1}{{x}_{0}+1}=\frac{\mathrm{log}(x+1)}{x}$

Since

${x}_{0}>0\Rightarrow \frac{1}{{x}_{0}+1}<1.$

$\Rightarrow 1>\frac{1}{{x}_{0}+1}=\frac{\mathrm{log}(x+1)}{x}\Rightarrow x>\mathrm{log}(x+1)$

I want to show that

$\frac{x}{1+x}<\mathrm{log}(1+x)<x$

for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.

$\frac{x}{1+x}<\mathrm{log}(1+x)\iff \frac{x}{1+x}-\mathrm{log}(1+x)<0$

Let

$f(x)=\frac{x}{1+x}-\mathrm{log}(1+x).$

Since

$f(0)=0$

and

${f}^{\prime}(x)=\frac{1}{(1+x{)}^{2}}-\frac{1}{1+x}<0$

for all $x>0$, $f(x)<0$ for all $x>0$. Is this correct so far?

I go on with the second part: Let

$f(x)=\mathrm{log}(x+1)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an ${x}_{0}$ between $a$ and $x$ with

${f}^{\prime}({x}_{0})=\frac{f(x)-f(a)}{x-a}\iff \frac{1}{{x}_{0}+1}=\frac{\mathrm{log}(x+1)}{x}$

Since

${x}_{0}>0\Rightarrow \frac{1}{{x}_{0}+1}<1.$

$\Rightarrow 1>\frac{1}{{x}_{0}+1}=\frac{\mathrm{log}(x+1)}{x}\Rightarrow x>\mathrm{log}(x+1)$

asked 2022-07-01

How solve this $\mathrm{log}{x}^{{\mathrm{log}}_{x}y}=\phantom{\rule{1em}{0ex}}\frac{5}{2}\phantom{\rule{0ex}{0ex}}x+y=6\phantom{\rule{0ex}{0ex}}$

How solve this logarithm equation?

How solve this logarithm equation?