We can use the Property of Equality for Logarithmic Functions to take the logarithm of both sides:

\(\displaystyle \small \log 3^x = \log 12\)

Use the Power Property of Logarithms:

\(\displaystyle \small x \log 3 = \log 12\)

Divide each side by \(\displaystyle \log3\) :

\(\displaystyle \small x=\frac{\log 12}{\log 3}\)

Use a calculator to get:

\(\displaystyle \small x \approx \frac{1.0792}{0.4771}\)   

or

\(\displaystyle \small x \approx 2.2620\)

Using Rules of Logarithm recall: 

\(\displaystyle \log a^b = b\log a\). 

Thus, in this situation we bring the 2 in front and we get our solution. 

\(\displaystyle 2 \log(6)\)

We can simplify the natural log exponents by using the following rules for naturla log.

\(\displaystyle \ln A+\ln B=\ln AB\)

\(\displaystyle \ln A - \ln B=\ln\frac{A}{B}\)

Using these rules, we can perform the following steps.

\(\displaystyle (e^{(\ln2+\ln93-\ln6)}e^{-1})^2\)

\(\displaystyle (e^{(\ln(2(93))-\ln6)}e^{-1})^2\)

\(\displaystyle (e^{(\ln(\frac{2(93)}{6})}e^{-1})^2\)

\(\displaystyle (e^{\ln 31}e^{-1})^2\)

Knowing that the e cancels the exponential natural log, we can cancel the first e.

\(\displaystyle (31e^{-1})^2\)

Distribute the square into the parentheses and calculate.

\(\displaystyle (31)^2(e^{-1})^2=961e^{-2}\)

Remember that a negative exponent is equivalent to a quotient. Write it as a quotient and then you're finished.

\(\displaystyle 961e^{-2}=\frac{961}{e^2}\)

Since the exponent is inside the parentheses, you must take the square of 1000 before finding the logarithim.  Therefore

\(\displaystyle \small \log(1000^{2})=\log (1000000)=6\)

because \(\displaystyle \small 10^{6}=1,000,000\)

Since the exponent is inside the parentheses, you must determine the value of the exponential expression first.

\(\displaystyle \small 9^{2}=81\)

then you solve the logarithm

\(\displaystyle \small \log _{3}(81)=4\)   because \(\displaystyle \small 3^{4}=81\)

\(\displaystyle \small \log _{a}(b)\) gives us the exponent to which \(\displaystyle \small a\) must be raised to yield \(\displaystyle \small b\)

When \(\displaystyle \small a\) is actually raised to that number in the equation given, the answer must be \(\displaystyle \small b\)

This is a simple exponent of a logarithmic answer.

\(\displaystyle \small \log(1000)=3\) because \(\displaystyle \small 10^{3}=1000\)

\(\displaystyle \small 3^{2}=9\)

This is a two step problem.  First find the log base 2 of 16

\(\displaystyle \small \log _{2}(16)=4\)   because \(\displaystyle \small 2^{4}=16\)

then 

\(\displaystyle \small 4^{3}=64\)

Since a logarithm answers the question of which exponent to raise the base to receive the number in parentheses, if the number in parentheses is the base raised to an exponent, the exponent must be the answer.

Before we do anything, the exponent of 4 must be moved to the front of the expression, as the rules of logarithms dictate. We end up with \(\displaystyle \small 4\ln{AB}\). Remember that a product inside of a logarithm can be rewritten as a sum: \(\displaystyle \small 4(\ln{A}+\ln{B})\). Distributing, we get \(\displaystyle \small 4\ln{A}+4\ln{B}\).