## Pre-Calculus Tutorial

*Exponential and Logarithmic Equations*

#### Intro

Exponential and Logarithmic equations are very simple; however, algebraic mistakes can easily be made. This problem illustrates how to find the solution set in terms of “e” and the solution set in decimal form.

#### Sample Problem

#### Solution

We are given 2ln(6x)=8

-The first thing the question asks is for the solution set in terms of “e”

-All that means is solve for x

2ln(6x)=8

2ln(6x)/2=8/2 -We divide by two on both sides

ln(6x)=4 -The 2 on the left side cancels out and on the right divides to 4

-our initial reaction is to divide the 6 on both sides, DO NOT! We have to think of the ‘ln’ and the ‘6x’ as married and in order to separate them we must get rid of the ‘ln’

-in order to get rid of the ‘ln’ we exponentiate both sides with base “e.”

Therefore,

e^ln(6x)= e^4

Then the “ln” and the “e” cancel out

6x=e^4

Now we may divide the 6 on both sides and we are left with

x=e^4/6 ==> and this is our solution set in terms of “e”

Our solution set in decimal form is simply the quotient of e^4/6 which is 9.10

# About The Author

Math And Science |

I am excellent in math with qualifications up to calculus thus far. I am great at English and composition. I am very qualified in sciences as well up to General Chemistry 1 thus far. I am decent at the social sciences but still qualified and a very well explainer. |