Which logarithmic equation is equivalent to the exponential equation below? E^a=28.37

2 answers:

7 0

1. To solve this problem, it is important to know that the logarithmic functions and the exponential functions are inverse. Then, you have:

e^a=28.37
 a=ln(28.37)

2. Therefore: Which logarithmic equation is equivalent to the exponential equation e^a=28.37? As you can see, the answer for this question is:

a=ln(28.37)

Travka [436]3 years ago

7 0

Answer:

ln 28.37= a

Step-by-step explanation:

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Oil leaks out of a tanker at a rate of r=f(t) liters per minute, where t is in minutes. If f(t) = A e^{-k t}, write a definite i

8_murik_8 [283]

Answer:

$V_{total} = \displaystyle\int_0^{60} A e^{-k t}~dt$

Step-by-step explanation:

We are given the following in the question:

Oil leaks out of a tanker at a rate of r = f(t) liters per minute, where t is in minutes.

$f(t) = A e^{-k t}$

Let V be the volume, then we are given that rate of leakage is:

$\displaystyle\frac{dV}{dt} = f(t) = A e^{-k t}$

Thus, we can write:

$dV = f(t).dt = A e^{-k t}~dt$

We have to find the a definite integral expressing the total quantity of oil which leaks out of the tanker in the first hour.

Thus, total amount of oil leaked will be the definite integral from 0 minutes to 60 minutes.

$dV =A e^{-k t}~dt\\V_{total} = \displaystyle\int_a^b f(t)~dt\\\\a = 0\text{ minutes}\\b = 60\text{ minutes}\\\\V_{total} = \displaystyle\int_0^{60} A e^{-k t}~dt$