SOLUTION: I am not even sure how to plug this in to my calculator. I do not know what order to take to break this down. Please help me solve for x. a. e^in(3-2x) =5x b. In x - In 2 = 1

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I am not even sure how to plug this in to my calculator. I do not know what order to take to break this down. Please help me solve for x. a. e^in(3-2x) =5x b. In x - In 2 = 1      Log On


   



Question 116560: I am not even sure how to plug this in to my calculator. I do not know what order to take to break this down. Please help me solve for x.
a. e^in(3-2x) =5x b. In x - In 2 = 1

Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
I am not even sure how to plug this in to my calculator.
I do not know what order to take to break this down.
Please help me solve for x.

a. e%5Eln%283-2x%29=5x b. ln%28x%29+-+ln%282%29+=+1

First of all what you thought was "In" was really "ln" which
means "natural logarithm" or loge where e = 2.718...

Secondly, you don't plug anything into a calculator.  The 
calculator is a nice tool for saving time, but it doesn't save 
you from having to learn the rules of logarithms, and they
are all that are required here:

For (a) and (b) you need to learn the rule of logarithms:

e%5Eln%28A%29=A

For (b) you need to learn the rule of logarithms:

ln%28A%29+-+ln%28B%29+=+ln%28A%2FB%29

--------------------
(a)
e%5Eln%283-2x%29=5x

Using the first rule above on the left side, we have

3-2x=5x

which you can easily solve and get

x=3%2F7

(b)
ln%28x%29+-+ln%282%29+=+1

Using the second rule above on the left side, we have

ln%28x%2F2%29+=+1

Now we raise e to the both sides power:

e%5Eln%28x%2F2%29+=+e%5E1

Using the first rule on the left side, and
erasing the 1 exponent on the right, we have:

x%2F2+=+e

or

x+=+2e

Edwin