Answer:
[tex]T_0=80695.17162...[/tex]
Explanation:
Given equation:
[tex]\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124[/tex]
To solve the given equation:
[tex]\textsf{Apply log rules}: \quad e^{\ln (x)}=x[/tex]
[tex]\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}[/tex]
Multiply both sides by T₀:
[tex]\implies T_0-100=T_0e^{-0.00124}[/tex]
Add 100 to both sides:
[tex]\implies T_0=T_0e^{-0.00124}+100[/tex]
Subtract [tex]T_0e^{-0.00124}[/tex] from both sides:
[tex]\implies T_0-T_0e^{-0.00124}=100[/tex]
Factor out the common term T₀:
[tex]\implies T_0(1-e^{-0.00124})=100[/tex]
Divide both sides by [tex](1-e^{-0.00124})[/tex]
[tex]\implies T_0=\dfrac{100}{1-e^{-0.00124}}[/tex]
Carry out the calculation:
[tex]\implies T_0=\dfrac{100}{1-0.99876...}[/tex]
[tex]\implies T_0=\dfrac{100}{0.001239231...}[/tex]
[tex]\implies T_0=80695.17162...[/tex]
