The first thing we notice about this problem is that $\displaystyle x$ is an exponent. This should be an immediate reminder: use logs!

The question is, which base should we choose for the log? We should use the natural log (log base e) because the right-hand side of the equation already has e as a base of an exponent. As you will see, things cancel out more nicely this way.

$\displaystyle 5^{x}=2e^5$

Take the natural log of both sides:

$\displaystyle \textup{ln}(5^x) = \textup{ln}(2e^5)$

Rewrite the right-hand side of the equation using the product rule for logs:

$\displaystyle \textup{ln}5^x = \textup{ln}2+\textup{ln}e^5$

Now rewrite the whole equation after bringing down those exponents.

$\displaystyle x\textup{ln}5 = \textup{ln}2+5\textup{ln}e$

$\displaystyle \textup{ln}e$ is the same thing as $\displaystyle \textup{log}_{e}e$, which equals 1.

$\displaystyle x\textup{ln}5 = \textup{ln}2+5$

Now we just divide by $\displaystyle \textup{ln}5$ on both sides to isolate $\displaystyle x$.

$\displaystyle x = \frac{\textup{ln}2+5}{\textup{ln}5}$

Using the properties of logarithms

$\displaystyle n \ln a = \ln a^{n}$ and $\displaystyle \ln a - \ln b = \ln \frac{a}{b}$,

we simplify as follows:

$\displaystyle 2 \ln x - \ln (x + 1)$

$\displaystyle = \ln x^{2} - \ln (x + 1)$

$\displaystyle = \ln \frac{x^{2} }{x + 1}$

By the reverse-FOIL method, we factor the polynomial as follows:

$\displaystyle x^{2} + 4x + 3 = (x + 1)(x + 3)$

Therefore, we can use the property 

$\displaystyle \ln ab = \ln a + \ln b$

as follows:

$\displaystyle \ln \left ( x^{2} + 4x + 3\right ) =\ln (x + 1)(x + 3) = \ln (x + 1) + \ln (x+3)$

The original equation is:

$\displaystyle 6e^{-x}+1=3$

Subtract $\displaystyle 1$ from both sides:

$\displaystyle 6e^{-x}=2$

Divde both sides by $\displaystyle 6$:

$\displaystyle e^{-x}=\frac{2}{6}=\frac{1}{3}$

Take the natural logarithm of both sides:

$\displaystyle -x=\ln \frac{1}{3}$

Divde both sides by $\displaystyle -1$ and use a calculator to get:

$\displaystyle x=-1.099$

Remember that $\displaystyle ln$ is still a logarithm of a positive number, $\displaystyle e$.

It's not possible to raise $\displaystyle e$ to ANY power and obtain a negative number. Because even $\displaystyle e^{-3}$, for example, is just $\displaystyle \frac{1}{e^3}$, which is a ratio of two positive numbers, and therefore positive.

More than that, it's also not possible to obtain 0 by raising $\displaystyle e$ to any power. Think: "To what power can I exponentiate e and obtain 0?"

So the domain is strictly positive. It excludes negative numbers and 0.

What about the range? To what possible values are we allowed to exponentiate $\displaystyle e$?

Well, we just saw that $\displaystyle e^x$ has a definition for negative numbers. $\displaystyle e^0 = 1$ (this fact is true for ALL numbers, not just $\displaystyle e$).

And we can obviously raise it to positive powers. So the range is all real numbers. It includes negative numbers, 0, and positive numbers.

Give both sides the same base, using e:

$\displaystyle e^{ln(x+3)}=e^3$.

Because e and ln cancel each other out, $\displaystyle x+3=e^3$.

Solve for x and round to the nearest tenth:

$\displaystyle x=e^3-3$

$\displaystyle x\approx17.1$

To solve for x, keep in mind that the natural logarithm and the exponential cancel each other out (property of any logarithm with a base that is being taken of that same base with an exponent attached). When they cancel, we are just left with the exponents:

$\displaystyle x+5=1$

$\displaystyle x=-4$

The natural log has a base of $\displaystyle e$.  This means that the term will simplify to whatever is the power of $\displaystyle e$.  Some examples are:

$\displaystyle ln(e^1) = 1$

$\displaystyle ln(e^2) = 2$

This means that $\displaystyle ln(e^{x^2-1}) = (x^2-1)$

Multiply this quantity with three.

$\displaystyle 3(x^2-1) = 3x^2-3$

The answer is:  $\displaystyle 3x^2-3$

In order to simplify this expression, use the following natural log rule.

$\displaystyle ln_x(x^y) =y$

The natural log has a default base of $\displaystyle e$. This means that: 

 $\displaystyle 2e\cdot ln(e^3)=2e\cdot ln_e(e^3) = 2e\cdot (3)$

The answer is:  $\displaystyle 6e$

According to log properties, the coefficient $\displaystyle 4$ in front of the natural log can be rewritten as the exponent raised by the quantity inside the log.

$\displaystyle 2e^{4ln{(2e)}} = 2e^{ln[{(2e)^4}]}$

Notice that natural log has a base of $\displaystyle e$.  This means that raising the log by base $\displaystyle e$ will eliminate both the $\displaystyle e$ and the natural log.

The terms become:  $\displaystyle 2(2e)^4$

Simplify the power.

$\displaystyle 2(2e)^4 = 2(16e^4) = 32e^4$

The answer is:  $\displaystyle 32e^4$