How to rearrange this equation

[michaelsmith559]

How can I rearrange this equation to solve for C169 and how can it be rearranged to solve for EXP(1)?

$B$2*(EXP(1)^(C169*A169))=D169

Thanks for the help.

Mike

 

[MickG]

Try:- EXP(1)=(D169/$B$2)^(1/(C169*A169)

 

[mikerickson]

EXP(1) is a constant. This gives you the relationship between the cells.

LN(D169)-LN($B$2) = C169*A169

 

[GooberTron]

(EXP(1)^(C169*A169)) doesn't make a lot of sense to me. This is surely the same as just doing EXP(C169*A169)?
If so then to just re-write in general algebra we have (where e is Euler's number);

b*e^ca=d

To solve for c just use natural logs;
ln(be^ca)=ln(d)
ln(b)+ln(e^ca)=ln(d)
ln(b)+ca*ln(e)=ln(d)
ln(b)+ca=ln(d) (as ln(e) = 1)
ca=ln(d)-ln(b)
ca=ln(d/b)
c=(ln(d/b))/a

To "solve" for e is a bit weird. Perhaps do the same but take logs to base 10 instead so you still have an "e" when you "anti-log".

 

[shg]

how can i rearrange this equation to solve for c169

=ln(d169 / $b$2) / a169

 

[michaelsmith559]

Works great. With this I can plug back in some values and see if I get Euler's number.

 

[michaelsmith559]

Let me ask this another way to clarify for myself. If I know $B$2, EXP(1) or 2.718281828, and A169 do I have to know D169 to solve for C169? Can I just use B2, Exp(1), and A169 to solve for C169?

 

[shg]

No.

Your equation in D169 can be (should be) rewritten as

=$B$2*EXP(C169*A169)

D169 clearly depends on C169. Solving for C169 clearly depends on the value of D169.

 

[michaelsmith559]

Okay. Is it correct to assume that trying to solve for C169 without D169 would be like trying to solve one equation with two unknowns, which is impossible?

 

[shg]

It's an equation in four variables. If you know any three, you can solve for the fourth.