i'm stuck on completing this equations. Is this correct? z=ae^(−bt) ln(z)=ln(a)+ln(e^(−bt)) ln(z)=ln(a)+(1)(−bt) ln(z)=ln(a)−bt

Ebone6v

Ebone6v

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2022-08-14

Exponential function to logarithmic function
i'm stuck on completing this equations. Is this correct?
z = a e b t
ln ( z ) = ln ( a ) + ln ( e b t )
ln ( z ) = ln ( a ) + ( 1 ) ( b t )
ln ( z ) = ln ( a ) b t

Answer & Explanation

Jaiden Gould

Jaiden Gould

Beginner2022-08-15Added 11 answers

It is correct!!!
If you want to solve for t, it is as followed:
ln ( z ) = ln ( a ) b t b t = ln ( a ) ln ( z ) = ln ( a z ) t = 1 b ln ( a z ) ,    where  z , a > 0
z = a e b t
if z = 0 a = 0  and  b , t  are arbitrary  .
polynnxu

polynnxu

Beginner2022-08-16Added 6 answers

Yes, you did apply well this property of logarithm: l n ( a b ) = l n ( a ) + l n ( b ), and the fact that l n ( x ) is the inverse of e x

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