antilogarithms

Question

In Algebra 2, I am given the following problem:

 

e^x = 6.8

 

I am told to solve using a scientific calculator. What is the answer to this equation?

Answer 1

x = ln 6.9

 

On your calculator locate the "ln" button.

Tap it.

Enter "6.8."

Tap "=."

1.91692261

Answer 2

Logarithms

 

y=log_b(x) if and only if x=b^y

ln(a^x)=x*ln(a)

 

If e^x=6.8, then we can use the natural log, LN or ln of both sides and the answer will be equal.

 

ln(e)=1 -> ln(e^x)=ln(6.8) 

 

Hence 

x=ln(6.8)

Answer 3

To solve this problem, we need to use some properties of logarithms. First, understand that a logarithm is equal to the the power to which you must raise a base in order to get some other number. For example, if you see log₂(4), it is the same thing as the power you need to raise the base (2) to in order to get 4. In this case, it is 2. (log₂(4)=2) Generally, you will be using the natural log, which is log_e(x), or just ln(x). Thus, ln(e) is the number you need to raise e to in order to get e (which is just 1, so ln(e)=1).

 

Now consider ln(e^x). This is asking what we need to raise e to in order to get e^x. Hopefully you can see that you simply need to raise e to the x power in order to get e^x. Thus, ln(e^x)=x.

 

So, going back to the original equation, we can simply take the natural log of both sides in order to get x alone on one side.

 

e^x=6.8 <=> ln(e^x)=ln(6.8)

 

We know that ln(e^x) = x, so we know that x=ln(6.8). You should be able to simply plug ln(6.8) into a scientific calculator. Doing this, we get

 

x=1.91692