Solving $e^x - 3 = 0$

Question

I want to solve this equation for $x$:

$$e^x - 3 = 0$$

Can somebody give me some hints?

Thanks.

Answer 1

First of all, add 3 to both sides, making $e^x = 3$. Next, use this link to solve $e^x = 3$: http://www.mathpapa.com/algebra-calculator.html

If you want an explanation, here you go:

$$e^x-3=0$$

Add $3$ to both sides:

$$e^x=3$$

Take the natural log of both sides:

$$\log(e^x) = \log(3)$$

The left side simplifies to:

$$x = \log(3)$$

Find the answer to that using a calculator.

Answer 2

$x=\ln(3)$.

Filling up the $30$ characters.

Answer 3

I'm assuming you don't know logarithms and you would like more precise explanation. First, bring the $-3$ to the other side: $$e^x=3$$ Now, to solve equations like this you need logarithms. Let's take a look at this: $2^3=8$ . Two raised to the third power is eight. Keep in mind, $2$ is the base. This equation is equal to $\log_28=3$. Notice that the base of the logarithm is $2$ as in $2^3=8$.

So, logarithms basically solve "to what power I have to raise a number, to get another number." In the equation $\log_28$ we can ask ourselves what number do I have to raise the base $(2)$ to get number $8$. It is $3$, because $2^3=8$.

The same stuff applies when the base is $e$. The equation $e^x=3$ is the same equation as $x=\log_e3$. So to what power do I have to raise $e$ to get $3$. $x=\log_e3$ is the same as $x=\ln3$, where $\ln$ means natural logarithm., or just logarithm with base $e$.

Now, normally you can leave your answer as $x=\ln3$, but if you want to get exact number that is eqal $x$, you just have to compute lnatural logarithm of $3$, which is irrational and it is approximately $1.0986122...$

So, $x=\ln3=1.0986122...$