Rearranging A Function with a Log Term

[MurraySchwartz]

I have the following issue:

The problem is:

10 = 1000 * ln(1000/x) Solve for x.

I tried rearranging to

10/1000 = ln(1000/x) but it does not work out.

Can anyone help with the proper way to do this?

 

[stapel]

10 = 1000 * ln(1000/x) Solve for x.

I tried rearranging to

10/1000 = ln(1000/x) but it does not work out.

What do you mean by "it does not work out"? You did the division on the left-hand side, you split the log on the right-hand side by applying a log rule (to get "ln(1,000) - ln(x)"), solved to get ln(x) by itself, converted the equation using exponentials, and... then what?

Please be complete. Thank you!

 

[MurraySchwartz]

What do you mean by "it does not work out"? You did the division on the left-hand side, you split the log on the right-hand side by applying a log rule (to get "ln(1,000) - ln(x)"), solved to get ln(x) by itself, converted the equation using exponentials, and... then what?

Please be complete. Thank you!

Thanks for your reply...Laying out what I have:

So now I have this:

10/1000 = ln(1000)-Ln(X) next,

10/1000 = 6.907755279 - ln(x) .leads to....

0.001 - 6.907755279 = ln(x) ...and finally...

SO - 6.897755279 = ln(x).....I know that's wrong?

 

[pka]

So now I have this:
10/1000 = ln(1000)-Ln(X) next,
10/1000 = 6.907755279 - ln(x) .leads to....
0.001 - 6.907755279 = ln(x) ...and finally...
SO - 6.897755279 = ln(x).....I know that's wrong?

\(\displaystyle \begin{align*} 10&=10^3\log(10^3)-10^3\log(x) \\\log(x) &= \log(10^3)-10^{-2}\end{align*}\)

Solve for \(\displaystyle x\).

 

[MurraySchwartz]

\(\displaystyle \begin{align*} 10&=10^3\log(10^3)-10^3\log(x) \\\log(x) &= \log(10^3)-10^{-2}\end{align*}\)

Solve for \(\displaystyle x\).

Ok..I think I have it.....

Log(x) = Log(10³)-10^-2

Log(x) = 6.907755279-0.01

Log(x) = 6.897755279

x = e^6.897755279

x = 990.0498337

Thank you very much!

 

[stapel]

10/1000 = ln(1000)-Ln(X) next,

10/1000 = 6.907755279 - ln(x) .leads to....

0.001 - 6.907755279 = ln(x) ...

What happened to the "minus" on the log on the right-hand side?

 

[MurraySchwartz]

What happened to the "minus" on the log on the right-hand side?

I divided both sides by 1000, then rearranged the equation.

 

[stapel]

I divided both sides by 1000, then rearranged the equation.

According to what you'd posted earlier, you'd converted the fraction on the left-hand side into decimal form, and then subtracted the six-point-whatever from the right-hand to the left-hand side. But—

What happened to the "minus" on the log?