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Delos · May 17, 2009, 09:04 AM

Can someone explain step by step how to make x the subject in an exponential equation?

I know the general form is y = ae^kx
So the just say I need to find out at what time would the experiment be at its half life which is 50, so I sub 50 into the following equation:

y = 92.776e^(-0.701t)
50=92.776e^(-0.701t)

I'm really bad at exponentials, could someone help? Step by step would be much appreciated :)

also another question,

lnY= mX + ln k

What is the step by step method to make Y the subject? :)
Thank you in advance...

DD

Unknown008 · May 17, 2009, 09:15 AM

Ok, here it goes:

$y=ae^{kx}$

Put log on both sides:

$ln(y)=ln(ae^{kx})$

$ln(y)=kxln(ae)$

$\frac{ln(y)}{kln(ae)}=x$

$\frac{ln(y)}{k(ln(a) + ln(e))}=x$

$\frac{ln(y)}{k(ln(a) +1)}=x$

However, I don't quite know what you want concerning the half life.

For the second one,

$lnY= mX + ln k$

$Y= e^{(mX + ln k)}$

It's the reverse of the previous process; here's the rule:

$a=b^c$

$log_b a = c$

Perito · May 17, 2009, 09:51 AM

[QUOTE]Can someone explain step by step how to make x the subject in an exponential equation?
/QUOTE]

Just remember that $Log(A^B) = B\, Log(A)$

That removes the exponent from the equation and makes it solvable by normal algebra. If it's in the form

$A^B=something$

you need to take the logarithm of both sides, first.