got it
Hi marksar
$\displaystyle T=\frac{175}{28}~\ln (1-\frac{V}{35})$
$\displaystyle \ln (1-\frac{V}{35})=\frac{28}{175}T$
$\displaystyle \log_e (1-\frac{V}{35}) = \frac{28}{175}T$
Hint :
$\displaystyle \log_a b = c \rightarrow b=a^c$
You can solve the rest questions using the same hint
Hi marksar
I don't think your answer is right. Continuing my work and using the hint given :
$\displaystyle \log_e (1-\frac{V}{35}) = \frac{28}{175}T$
$\displaystyle 1-\frac{V}{35}=e^{\frac{28}{175}T}$
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