Respuesta :
if I read you correctly with [tex]\bf -a=b^{\frac{}{}^\frac{-c}{t^2}}[/tex]
then
[tex]\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)}[/tex]
now, without using the change of base rule, you can also just leave it as [tex]\bf t=\sqrt{\cfrac{-c\cdot ln(b)}{ln(-a)}} [/tex]
then
[tex]\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)}[/tex]
now, without using the change of base rule, you can also just leave it as [tex]\bf t=\sqrt{\cfrac{-c\cdot ln(b)}{ln(-a)}} [/tex]
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Isolate T in the formula:
[tex]\mathsf{-a=b\cdot e^{-\,C/T^2}}[/tex]
Assuming b ≠ 0, you can write
[tex]\mathsf{-\,\dfrac{a}{b}=e^{-\,C/T^2}}[/tex]
Now, take the natural logarithm of both sides, and you get
[tex]\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)}}[/tex]
Taking the square root of both sides, you have
[tex]\mathsf{T=\pm\,\sqrt{\dfrac{-\,C}{\ell n\!\left(-\frac{a}{b}\right)}}}[/tex]
So, there are two possibilities for T:
[tex]\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}[/tex]
I hope this helps. =)
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Isolate T in the formula:
[tex]\mathsf{-a=b\cdot e^{-\,C/T^2}}[/tex]
Assuming b ≠ 0, you can write
[tex]\mathsf{-\,\dfrac{a}{b}=e^{-\,C/T^2}}[/tex]
Now, take the natural logarithm of both sides, and you get
[tex]\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)}}[/tex]
Taking the square root of both sides, you have
[tex]\mathsf{T=\pm\,\sqrt{\dfrac{-\,C}{\ell n\!\left(-\frac{a}{b}\right)}}}[/tex]
So, there are two possibilities for T:
[tex]\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}[/tex]
I hope this helps. =)
