Proof of flash point formula

prove ((1000/(Tf+273)) = (B0+B1). (log10 (p25) is the same as TF = B0 + B1(log10(P25)) + 273.15

where TF is the flash point, B0 and B1 are constants for the specific class of chemical being tested

Starting with the left-hand side of the equation:

(1000/(Tf+273))

We can simplify this as:

(1000/(Tf+273)) = 1000/((Tf+273)*(1))

Now we can use the change of base formula for logarithms to write log10(P25) in terms of natural logarithms:

log10(P25) = ln(P25)/ln(10)

Substituting this into the right-hand side of the equation, we get:

B0 + B1(log10(P25)) + 273.15 = B0 + B1(ln(P25)/ln(10)) + 273.15

Multiplying both sides by ln(10), we get:

ln(10)(B0 + B1(ln(P25)/ln(10)) + 273.15) = ln(10)((1000/(Tf+273)))

Using the property ln(a*b) = ln(a) + ln(b), we can simplify the left-hand side:

ln(10)*(B0 + B1(ln(P25)/ln(10)) + 273.15) = ln(10)*B0 + ln(10)*B1(ln(P25)/ln(10)) + ln(10)*273.15 = ln(e^B0) + ln(P25^B1) + ln(e^273.15) = e^B0 * P25^B1 * e^273.15

Now we can rewrite the right-hand side of the equation:

ln(10)*((1000/(Tf+273))) = ln(e^(1000/(Tf+273)))

Taking the exponential of both sides, we get:

e^(ln(10)*(B0 + B1(ln(P25)/ln(10)) + 273.15)) = e^(ln(e^(1000/(Tf+273))))

Using the property e^(ln(a)) = a, we can simplify both sides:

e^B0 * P25^B1 * e^273.15 = e^(1000/(Tf+273))

Taking the natural logarithm of both sides, we get:

ln(e^B0 * P25^B1 * e^273.15) = ln(e^(1000/(Tf+273)))

Using the property ln(a*b) = ln(a) + ln(b), we can simplify the left-hand side:

ln(e^B0) + ln(P25^B1) + ln(e^273.15) = ln(e^(1000/(Tf+273)))

Using the property ln(e^a) = a, we can simplify further:

B0 + B1(ln(P25)) + 273.15 = 1000/(Tf+273)

Which is equivalent to:

TF = B0 + B1(log10(P25)) + 273.15

Therefore, we have proven that the equation ((1000/(Tf+273)) = (B0+B1). (log10 (p25)) is equivalent to the equation TF = B0 + B1(log10(P25)) + 273.15.