   Chapter 4.6, Problem 3SWU ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views

# In Exercises 1-4, solve the equation for k. 25 = 16 e − 0.01 k

To determine

To calculate: The value of k in the equation 25=16e0.01k.

Explanation

Given Information:

The provided equation is 25=16e0.01k.

Formula used:

If f(x)=ex is a function then the natural logarithm of the function that is f(x)=lnex is given by the inverse property that is,

lnex=x

If the function is f(x)=lnab, then the property of natural logarithm is given by.

lnab=lnalnb

Calculation:

Consider the equation, 25=16e0.01k

Divide both sides by 16.

2516=1616e0.01k2516=e0.01k

Take natural logarithm each side

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