The expression can be rewritten as a fraction.

\(\displaystyle log_{100}10 = \frac{log(10)}{log(100)}\)

The log based 100 can be rewritten as ten squared.  

\(\displaystyle \frac{log(10)}{log(100)} = \frac{log(10)}{log(10^2)} = \frac{log(10)}{2log(10)} = \frac{1}{2}\)

The answer is:  \(\displaystyle \frac{1}{2}\)

For logs that are added, the inner quantities can be multiplied with each other to be combined as a single log.

For logs that are subtracted, the inner quantities would each be divided instead.

Rewrite the expression:

\(\displaystyle log(2)+log(4)+log(6)-log(8) =log( \frac{2(4)(6)}{8})\)

\(\displaystyle = log(6)= log(2)+log(3)\)

The answer is:  \(\displaystyle log(2)+log(3)\)

Recall that log has a base of ten.

The equation can be rewritten as:

\(\displaystyle log_{10}(\frac{x}{10})= 2\)

\(\displaystyle log_{10}(x)-log_{10}(10) = 2\)

\(\displaystyle log_{10}(x)-1= 2\)

Add 1 on both sides.

\(\displaystyle log_{10}(x)-1+1= 2+1\)

\(\displaystyle log_{10}(x) = 3\)

To eliminate the log based 10 on the left, we will need to raise both terms as exponents of base 10.

\(\displaystyle 10^{log_{10}(x)} = 10^3\)

\(\displaystyle x=1000\)

The answer is:  \(\displaystyle 1000\)

First, we start with the full equation:

\(\displaystyle log_{3}(x)=4\)

Now we can expand the right side of the equation:

\(\displaystyle log_{3}(x)=1+1+1+1\)

A log of it's own base equals \(\displaystyle 1\):

\(\displaystyle log_{3}(x)=log_{3}(3)+log_{3}(3)+log_{3}(3)+log_{3}(3)\)

Now we add the logs on the right side of the equation by multiplying the terms inside the logs:

\(\displaystyle log_{3}(x)=log_{3}(3*3*3*3)\)

\(\displaystyle log_{3}(x)=log_{3}(81)\)

\(\displaystyle x=81\)

By definition, a logarithm of any base that has the term \(\displaystyle 1\) inside is equal to \(\displaystyle 0\).  So we set that term equal to \(\displaystyle 1\):

\(\displaystyle x-4=1\)

\(\displaystyle x=5\)

When a logarithm equals \(\displaystyle 1\), the equation in the logarithm equals the logarithms base:

\(\displaystyle 2x=5\)

\(\displaystyle x=\frac{5}{2}\)

Rearranging the logarithm so that we exchange an exponent for the log we get:

\(\displaystyle 10^{-1}=x\)

\(\displaystyle x=.1\)

First we rearrange the equation, trading the logarithm for an exponent:

\(\displaystyle 8^1=2x+4\)

And then we solve:

\(\displaystyle 4=2x\)

\(\displaystyle 2=x\)

The first thing we can do is combine the log terms:

\(\displaystyle log_{3}((x+1)x)=2\)

Now we can change to exponent form:

\(\displaystyle 3^2=(x+1)x\)

We can combine terms and set the equation equal to \(\displaystyle 0\) to have a quadratic equation:

\(\displaystyle x^2+x-6=0\)

We then solve the equation and get the answers:

\(\displaystyle 2\) and \(\displaystyle -3\)

\(\displaystyle -3\) can't be an answer, because the values inside a log can't be negative, so that leaves us with a single answer of \(\displaystyle 2\).

The first thing we can do is combine log terms:

\(\displaystyle log_2((x+1)(x-1))=3\)

Simplifying the log term gives:

\(\displaystyle log_2(x^2-1)=3\)

Now we can change the equation to exponent form:

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

And to solve:

\(\displaystyle x^2-1=8\)

\(\displaystyle x^2=9\)

\(\displaystyle x=3, x-3\)

Here, the solution can't be \(\displaystyle -3\) because the term inside a logarithm can't be negative, so the only solution is \(\displaystyle 3\).