If the $\mathrm{log}(x)=-7.65$, how do you find x?

Damien Horton
2022-07-28
Answered

If the $\mathrm{log}(x)=-7.65$, how do you find x?

You can still ask an expert for help

Shelby Strong

Answered 2022-07-29
Author has **9** answers

remember: $\mathrm{log}(x)$ means ${\mathrm{log}}_{10}(x)$

log property: ${a}^{{\mathrm{log}}_{10}(x)}=x$

${\mathrm{log}}_{10}(x)=-7.65$

${10}^{{\mathrm{log}}_{10}(x)}={10}^{-7.65}$

$x={10}^{-7.65}$

x=0.000000023872 or $2.23872\times {10}^{-8}$

log property: ${a}^{{\mathrm{log}}_{10}(x)}=x$

${\mathrm{log}}_{10}(x)=-7.65$

${10}^{{\mathrm{log}}_{10}(x)}={10}^{-7.65}$

$x={10}^{-7.65}$

x=0.000000023872 or $2.23872\times {10}^{-8}$

Makena Preston

Answered 2022-07-30
Author has **3** answers

$1.\mathrm{log}(x)=-7.65$

$2.{10}^{\mathrm{log}(x)}={10}^{-7.65}$ ---> when 10 to the powerof log, log and 10 cancel each other.

$3.X={10}^{-7.65}$

$2.{10}^{\mathrm{log}(x)}={10}^{-7.65}$ ---> when 10 to the powerof log, log and 10 cancel each other.

$3.X={10}^{-7.65}$

asked 2022-11-06

Solving logarithmic equations including x

Let

$${\mathrm{log}}_{3}(x-2)=6-x$$

It's obvious drawing the graphs of the two functions that the only solution is $x=5$. But this is not really a proof, rather than observation.

How do you prove it algebraically?

Let

$${\mathrm{log}}_{3}(x-2)=6-x$$

It's obvious drawing the graphs of the two functions that the only solution is $x=5$. But this is not really a proof, rather than observation.

How do you prove it algebraically?

asked 2022-06-22

Given the exponential equation ${4}^{x}=64$, what is the logarithmic form of the equation in base $10$?

Would it be $\frac{{\mathrm{log}}_{10}64}{{\mathrm{log}}_{10}4}$?

Would it be $\frac{{\mathrm{log}}_{10}64}{{\mathrm{log}}_{10}4}$?

asked 2022-06-01

Is there a proof that ${n}^{x}{m}^{x}=({n}^{x}{)}^{(\mathrm{log}(mn)/\mathrm{log}(n))}$?

This isn't a homework question, just something I'm curious about, but you can treat it that way if you like.

So the other day I was playing with my calculator and I noticed that

${2}^{x}{10}^{x}=({2}^{x}{)}^{(\mathrm{log}(20)/\mathrm{log}(2))}$

I tried it out with some other numbers and came to the conclusion that

${n}^{x}{m}^{x}=({n}^{x}{)}^{(\mathrm{log}(nm)/\mathrm{log}(n))}$

So I wanted to see if there is a way to prove that.

I already know that $m={n}^{(\mathrm{log}(m)/\mathrm{log}(n))}$ and I figured that there must be a relation. So from that I can see that $mn={n}^{(\mathrm{log}(nm)/\mathrm{log}(n))}$. However I don't understand why that would mean that $(nm{)}^{x}=({n}^{x}{)}^{\mathrm{log}(nm)/\mathrm{log}(n)}$.

Is what I say actually true? How do the powers fit into the proof?

This isn't a homework question, just something I'm curious about, but you can treat it that way if you like.

So the other day I was playing with my calculator and I noticed that

${2}^{x}{10}^{x}=({2}^{x}{)}^{(\mathrm{log}(20)/\mathrm{log}(2))}$

I tried it out with some other numbers and came to the conclusion that

${n}^{x}{m}^{x}=({n}^{x}{)}^{(\mathrm{log}(nm)/\mathrm{log}(n))}$

So I wanted to see if there is a way to prove that.

I already know that $m={n}^{(\mathrm{log}(m)/\mathrm{log}(n))}$ and I figured that there must be a relation. So from that I can see that $mn={n}^{(\mathrm{log}(nm)/\mathrm{log}(n))}$. However I don't understand why that would mean that $(nm{)}^{x}=({n}^{x}{)}^{\mathrm{log}(nm)/\mathrm{log}(n)}$.

Is what I say actually true? How do the powers fit into the proof?

asked 2021-11-08

Use the properties of logarithms to expand the following expression as much as possible. Simplify any numerical expressions that can be evaluated without a calculator.

${\mathrm{log}}_{4}(48x+16y)$

asked 2022-10-09

Logarithmic equation help

${\mathrm{log}}_{5}(x+3)+{\mathrm{log}}_{5}(3x-5)={\mathrm{log}}_{25}\left(9{x}^{2}\right)$

I have the answer: $\left\{\frac{\sqrt{181}-1}{6}\right\}$ (only answer that falls in the domain)

i understand how to work with the left side but can't figure out how to get the RH side to base 5.

${\mathrm{log}}_{5}(x+3)+{\mathrm{log}}_{5}(3x-5)={\mathrm{log}}_{25}\left(9{x}^{2}\right)$

I have the answer: $\left\{\frac{\sqrt{181}-1}{6}\right\}$ (only answer that falls in the domain)

i understand how to work with the left side but can't figure out how to get the RH side to base 5.

asked 2022-10-28

Why is $$-\mathrm{ln}x$$ is not equal to $$1/\mathrm{ln}x$$?

I am doing differential equation now and I need to convert them into the proper form in order to do my homogeneous differential equation. So now I just found out that $$-\mathrm{ln}x$$ is not equal to $$1/\mathrm{ln}x$$. I thought it should be able to convert to $$-\mathrm{ln}x$$ to the negative 1 then I can put it into the form $$1/\mathrm{ln}x$$. Can anyone explain about it?

I am doing differential equation now and I need to convert them into the proper form in order to do my homogeneous differential equation. So now I just found out that $$-\mathrm{ln}x$$ is not equal to $$1/\mathrm{ln}x$$. I thought it should be able to convert to $$-\mathrm{ln}x$$ to the negative 1 then I can put it into the form $$1/\mathrm{ln}x$$. Can anyone explain about it?

asked 2022-01-21

Prove by Mean Value Theorem $\frac{x}{1+x}<\mathrm{ln}(1+x)<x\text{}\text{for}\text{}x0$