How can we write log _n (v / v₀) = 2x as v = v₀e^2x ? Please explain.

Dear Manish Sahu,

This is a rule of logarithm that 1/ln (x) becomes e^x .

This is the same thing that has happened here .

Log_n has gone as a denominator to the other side making it exponential !!

Hope this helped you immensely..

All the Very Best & Good Luck to you ...

Regards,

AskIITians Expert,

Godfrey Classic Prince

IIT-Madras

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log v - log vo = 2x (log to base n)

lnv - lnvo = 2x ln (n)

ln (v/vo) = 2x ln(n)

e^2xln(n) = v/vo