How Do I Correctly Exponentiate in Logarithmic Equations?

[vorcil]

I solved for the equation

\frac{f}{2} ln\frac{Tf}{Ti} = - ln \frac{vf}{vi}

I know to simplify the equation I need to Exponentiate both sides

<br /> <br /> e^{\frac{f}{2}} * e^{ln\frac{Tf}{Ti}} = e^{ln\frac{vi}{vf}}

note i changed - ln \frac{vf}{vi} \textrm{ too } ln\frac{vi}{vf} and removed the negative sign

,

after exponentiating both sides I get

e^{\frac{f}{2}} * \frac{Tf}{Ti} = \frac{Vi}{Vf}

But I'm not sure if I put the f/2 in the right place

My equation should be reading
{\frac{Tf}{Ti}}^\frac{f}{2}
but I don't know how to get to that
from exponentiating \frac{f}{2} \frac{Tf}{Ti}

could someone please expain the rules behind this?

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original question:

solve an equation for the work done in adiabatic compression

-

U = \frac{f}{2} N \kappa T

dU = \frac{f}{2} N \kappa dT

the work done during quasistatic compression is -P dv

\frac{f}{2} N\kappa dT = -P dV

using the ideal gas law to write P the pressure in terms of the variables T and v
ideal gas law: \frac{N \kappa T}{V} = P

i get

\frac{f}{2} N\kappa dT = -\frac{N\kappa T}{V} dv

simplifying by canceling out N K gives

\frac{f}{2} dT = -\frac{T}{V}dv

separation of variables because this is a differential equation

<br /> \frac{f}{2} \frac{dT}{T} = -\frac{dV}{V}

integrating both sides with respect to initial and final volume/temperatures
I get the equation

\frac{f}{2} \int_{Ti}^{Tf} \frac{1}{T}dT = -\int_{Vi}^{Vf} \frac{1}{V}dV

integrating

\left \frac{f}{2} lnT \right|_{Ti}^{Tf}= \left -lnV\right|_{Vi}^{Vf}

evaluating

\frac{f}{2} lnTf - lnTi = -lnVf + lnVi

noting that lnx - lny is equivalent to ln(x/y)

simplifying to get

\frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf}

this is where I was having the exponential problem

exponentiating both sides

do i raise e to the whole of the left side? or the individual parts?
e^\frac{f}{2} e^{ln\frac{Tf}{Ti}} \textrm{ or do i } e^{\frac{f}{2} ln{Tf}{Ti}
what happens to the f/2?
I know each of Tf and Ti should be raised to the power of f/2 afterwards

but I do not understand how that works

 

[Je m'appelle]

\frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf}

this is where I was having the exponential problem

exponentiating both sides

do i raise e to the whole of the left side? or the individual parts?
e^\frac{f}{2} e^{ln\frac{Tf}{Ti}} \textrm{ or do i } e^{\frac{f}{2} ln{Tf}{Ti}
what happens to the f/2?
I know each of Tf and Ti should be raised to the power of f/2 afterwards

but I do not understand how that works

Quick info about exponentials:

x^ax^b = x^{(a+b)}

\frac{x^a}{x^b} = x^{(a-b)}

(x^a)^b = x^{ab}

So, when you exponentiate

\frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf}

You get

e^{\frac{f}{2} ln\frac{T_f}{T_i}} = e^{ln\frac{V_i}{V_f}}

But remember that

\frac{f}{2} ln\frac{T_f}{T_i} = ln(\frac{T_f}{T_i})^{(\frac{f}{2})}

Plug-in that on your equation to get

(\frac{T_f}{T_i})^{(\frac{f}{2})} = \frac{V_i}{V_f}

Was that what you wanted?

________________________

EDIT:

Instead, if you had done

e^{\frac{f}{2}}e^{ln\frac{T_f}{T_i}} = e^{ln\frac{V_i}{V_f}}

On the left-hand side, you would be implying that

e^{\frac{f}{2}}e^{ln\frac{T_f}{T_i}} = e^{(\frac{f}{2}+ln\frac{T_f}{T_i})}

Which is incorrect, as you can clearly see, because the original expression was

\frac{f}{2}ln\frac{T_f}{T_i}

And not

\frac{f}{2}+ln\frac{T_f}{T_i}

So a quick explanation to your question would be reviewing the first exponential property that I've posted above, that is, the product one.