Answered

Isolate for T in -a=bexp(-C/T^2)

What I think is the answer is posted below. \(T= \sqrt{ \frac{-C}{ln( \frac{-a}{b} } }\)

Verified Answer :

if I read you correctly with \(\bf -a=b^{\frac{}{}^\frac{-c}{t^2}}\)

then

\(\bf -a=b^{\frac{}{}^\frac{-c}{t^2}}\implies ln(-a)=ln\left(b^{\frac{}{}^\frac{-c}{t^2}} \right)\implies ln(-a)=\cfrac{-c}{t^2}ln(b) \\\\\\ ln(-a)=\cfrac{-c\cdot ln(b)}{t^2}\implies t^2=\cfrac{-c\cdot ln(b)}{ln(-a)} \implies t^2=-c\cdot \cfrac{ln(b)}{ln(-a)} \\\\\\ \textit{recall the change of base rule}\to log_{{ a}}{{ b}}\implies \cfrac{log_{{ c}}{{ b}}}{log_{{ c}}{{ a}}}\qquad thus \\\\\\ t^2=-c\cdot ln_{-a}(b)\implies t=\sqrt{-c\cdot ln_{-a}(b)}\)

now, without using the change of base rule,  you can also just leave it as \(\bf t=\sqrt{\cfrac{-c\cdot ln(b)}{ln(-a)}}\)

lukyo

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Isolate T in the formula:

\(\mathsf{-a=b\cdot e^{-\,C/T^2}}\)


Assuming b ≠ 0, you can write

\(\mathsf{-\,\dfrac{a}{b}=e^{-\,C/T^2}}\)


Now, take the natural logarithm of both sides, and you get

\(\mathsf{\ell n\!\left(-\,\dfrac{a}{b}\right)=\ell n\big(e^{-\,C/T^2}\big)}\\\\\\ \mathsf{\ell n\!\left(-\,\dfrac{a}{b}\right)=-\,\dfrac{C}{T^2}}\\\\\\ \mathsf{T^2\cdot \ell n\!\left(-\,\dfrac{a}{b}\right)=-\,C}\\\\\\ \mathsf{T^2=\dfrac{-\,C}{\ell n\!\left(-\,\frac{a}{b}\right)}}\)


Taking the square root of both sides, you have

\(\mathsf{T=\pm\,\sqrt{\dfrac{-\,C}{\ell n\!\left(-\frac{a}{b}\right)}}}\)


So, there are two possibilities for T:

\(\mathsf{T=-\,\sqrt{\dfrac{-\,C}{\ell n\!\left(-\frac{a}{b}\right)}}~~~or~~~T=\sqrt{\dfrac{-\,C}{\ell n\!\left(-\frac{a}{b}\right)}}\qquad\qquad\checkmark}\)


I hope this helps. =)