# Help with rearranging formula

• Jun 11th 2009, 10:41 AM
Tino1210
Help with rearranging formula
Hello, Im looking for some guidance in rearranging formula. Need to ask if what I'm doing is right?

1) 4 (^x+1) = 6 ^0.5x , find x
ln4 (^x+1) = ln6(^0.5x)
x+1(ln4) = 0.5xln6
1.38x +1.38 = 0.895x
1.38 = (0.895/1.38) x
1.38 = 0.65x
x= 0.65/1.38
???

2) I = Imax (1-e ^-Rt/L) , R=30, L=0.01 - find t in seconds for current to reach 95%

= I/Imax = 1- e -Rt/1
=I/Imax+1 = -e -Rt/1
-ln (I/Imax+1) = -Rt/1 lne
-Rt/1= -ln(I/Imax+1)
-Rt= -ln(I/Imax +1) (l)
t= -ln (1/0.95 +1) (0.01)/-30
???

3) PV^Y = K , make Y the subject
V^Y = K/P
lnV^Y = ln K/P
YlnV = ln K/P
Y = K/P / lnV

4) V= Vmax exp -R/1 t , make t the subject
V/Vmax = e -R/1 t
ln V/Vmax = ln(e- R/1 t)
ln V/Vmax = -R/1 t lne
ln (V/Vmax) = -R/1 t
t = ln (V/Vmax) L/ -R
• Jun 11th 2009, 12:00 PM
Amer
Quote:

Originally Posted by Tino1210
Hello, Im looking for some guidance in rearranging formula. Need to ask if what I'm doing is right?

1) 4 (^x+1) = 6 ^0.5x , find x
ln4 (^x+1) = ln6(^0.5x)
x+1(ln4) = 0.5xln6
1.38x +1.38 = 0.895x
1.38 = (0.895/1.38) x ....
1.38 = 0.65x
x= 0.65/1.38
???

2) I = Imax (1-e ^-Rt/L) , R=30, L=0.01 - find t in seconds for current to reach 95%

= I/Imax = 1- e^ -Rt/1
=I/Imax+1 = -e^ -Rt/1
-ln (I/Imax+1) = -Rt/1 lne
-Rt/1= -ln(I/Imax+1)
-Rt= -ln(I/Imax +1) (l)
t= -ln (1/0.95 +1) (0.01)/-30
???

3) PV^Y = K , make Y the subject
V^Y = K/P
lnV^Y = ln K/P
YlnV = ln K/P
Y = ln(K/P) / lnV just you forget blue ln

4) V= Vmax exp -R/1 t , make t the subject
V/Vmax = e -R/1 t
ln V/Vmax = ln(e- R/1 t)
ln V/Vmax = -R/1 t lne
ln (V/Vmax) = -R/1 t
t = ln (V/Vmax) L/ -R

1)what you did here
$1.38x+1.38=0.895x$

$
\Rightarrow 1.38=.895x-1.38x\Rightarrow1.38=-.935x$

$
\Rightarrow x=\frac{-1.38}{.935}$

2) $I/Imax = 1- e^ -Rt/1$

$
\Rightarrow \frac{I}{Imax}-1 = -e^{\frac{-Rt}{L}}
$

$
\Rightarrow 1-\frac{I}{Imax}=e^{\frac{-Rt}{L}}$

$
\Rightarrow ln(1-\frac{I}{Imax})=$
$\frac{-Rt}{L}$

$
\Rightarrow \frac{(L)ln(1-\frac{I}{Imax})}{-R}=t$

the last one is not clear
• Jun 11th 2009, 12:05 PM
Tino1210
thanks for that, yeah the last one is hard to type out, sorry
• Jun 11th 2009, 12:08 PM
Amer
what I can't understand is e to power what this is what you ask about

$V=(Vmax)e^{\frac{-R}{L}} t$
• Jun 11th 2009, 12:09 PM
Tino1210
Quote:

Originally Posted by Amer
what I can't understand is e to power what this is what you ask about

$V=(Vmax)e^{\frac{-R}{L}} t$

yes, the t is positioned in the question between the -R/L
• Jun 11th 2009, 12:23 PM
Amer
Quote:

Originally Posted by Amer
what I can't understand is e to power what this is what you ask about

$V=(Vmax)e^{\frac{-R}{L}} t$

$V=(Vmax)e^{\frac{-R}{L}}t$

$\frac{V}{Vmax}=e^{\frac{-R}{L}}t$

$t=\frac{V}{V(e^{\frac{-R}{L}})}$

• Jun 11th 2009, 12:27 PM
Tino1210
yeah thats it, thanks
• Jun 11th 2009, 12:31 PM
Tino1210
is this one correct:

D = Do + log (x/y) , make x the subject

D- Do = log (x/y)
ln (D-Do) = x/y
y ln (D-Do) = x
• Jun 11th 2009, 12:35 PM
Amer
Quote:

Originally Posted by Tino1210
is this one correct:

D = Do + log (x/y) , make x the subject

D- Do = log (x/y)
ln (D-Do) = x/y
y ln (D-Do) = x

no it is wrong

you know that

$log A = log_{10} (A)$ if you do not know that now you know it

$D-Do = log(x/y)$

$10^{(D-Do)} = \frac{x}{y}$

$y(10^{(D-Do)}) = x$