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transposition of formulae involving logs and exponentials

hi could someone assist with this:

n=R + 10 ln p2p1\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 p2p1\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?


Original post by hitch1983
hi could someone assist with this


N=R + 10 ln p2p1\frac{p2}{p1}
N-R= 10 ln p2p1\frac{p2}{p1}
N−R10\frac{N-R}{10} = ln p2p1\frac{p2}{p1}

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

p2p1\frac{p2}{p1} = e N−R10\frac{N-R}{10}
(as in e to the power of N−R10\frac{N-R}{10})

p1 = p2 X e - (N−R10\frac{N-R}{10})

It won't let me write a fraction within a fraction..
(edited 9 years ago)
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Avatar for hitch1983
hitch1983
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thanks for the help
Original post by Complex solution
N=R + 10 ln p2p1\frac{p2}{p1}
N-R= 10 ln p2p1\frac{p2}{p1}
N−R10\frac{N-R}{10} = ln p2p1\frac{p2}{p1}

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

p2p1\frac{p2}{p1} = e N−R10\frac{N-R}{10}
(as in e to the power of N−R10\frac{N-R}{10})

p1 = p2 X e - (N−R10\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? p2e−((N−R)10)p_2\text{e}^{-\left(\frac{(N-R)}{10}\right)} or did you want to write p1=p2e(N−R10)p_1=\frac{p_2}{\text{e}^{\left( \frac{N-R}{10}\right)}}
(edited 9 years ago)
Original post by brianeverit
What do you mean by "a fraction within a fraction"? Do you mean like this? p2e−((N−R)10)p_2\text{e}^{-\left(\frac{(N-R)}{10}\right)} or did you want to write p1=p2e(N−R10)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.
(edited 9 years ago)
Original post by arkanm


Thanks :biggrin::biggrin:

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