transposition of formulae involving logs and exponentials

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hitch1983
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#1
Report Thread starter 6 years ago
#1
hi could someone assist with this:

n=R + 10 ln \frac{p2}{p1}

i have to change the subject to p1

this is my working

R goes over to the other side and subtracts with with n

R-n =10 ln
\frac{p2}{p1}

then i remove the fraction and take p2 over and multiply it

p2 (R-n) = 10 ln P1

take the e 10 on both sides by dividing it

p2 (R-n)
----------
E 10

problem solved

is this correct?


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Complex solution
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(Original post by hitch1983)
hi could someone assist with this
N=R + 10 ln \frac{p2}{p1}
N-R= 10 ln \frac{p2}{p1}
\frac{N-R}{10} = ln \frac{p2}{p1}

so...now just use the fact that a = e ln(a)

\frac{p2}{p1} = e \frac{N-R}{10}
(as in e to the power of \frac{N-R}{10})

p1 = p2 X
e
- (\frac{N-R}{10})

It won't let me write a fraction within a fraction..
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hitch1983
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thanks for the help
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brianeverit
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(Original post by Complex solution)
N=R + 10 ln \frac{p2}{p1}
N-R= 10 ln \frac{p2}{p1}
\frac{N-R}{10} = ln \frac{p2}{p1}

so...now just use the fact that a = e ln(a)

\frac{p2}{p1} = e \frac{N-R}{10}
(as in e to the power of \frac{N-R}{10})

p1 = p2 X
e
- (\frac{N-R}{10})

It won't let me write a fraction within a fraction..
What do you mean by "a fraction within a fraction"? Do you mean like this? p_2\text{e}^{-\left(\frac{(N-R)}{10}\right)} or did you want to write p_1=\frac{p_2}{\text{e}^{\left( \frac{N-R}{10}\right)}}
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Complex solution
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(Original post by brianeverit)
What do you mean by "a fraction within a fraction"? Do you mean like this? p_2\text{e}^{-\left(\frac{(N-R)}{10}\right)} or did you want to write p_1=\frac{p_2}{\text{e}^{\left( \frac{N-R}{10}\right)}}
(Original post by arkanm)
Probably the latter, hence "fraction within a fraction"
Yes, 'twas the latter.

I wish there was a tutorial for writing maths equations in forums.
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Complex solution
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Report 6 years ago
#6
Thanks
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