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 \(\displaystyle 3\).

First, let's change the equation to exponent form:

\(\displaystyle x^2+7=2^5\)

Then simplify:

\(\displaystyle x^2+7=32\)

And solve:

\(\displaystyle x^2=25\)

\(\displaystyle x=5, x=-5\)

Both answers are valid because \(\displaystyle x\) in the original equation is squared, so any negative numbers don't cause the logarithm to become negative.

We can start by getting both the log terms on the same side of the equation:

\(\displaystyle log_6(4x)-log_6(x+9)=0\)

Then we combine log terms:

\(\displaystyle log_6(\frac{4x}{x+9})=0\,\)

Now we can change to exponent form:

\(\displaystyle \frac{4x}{x+9}=6^0\)

Anything raised to the \(\displaystyle 0\)th power equals \(\displaystyle 1\), and from here it becomes a simpler problem to solve:

\(\displaystyle \frac{4x}{x+9}=1\)

\(\displaystyle 4x=x+9\)

\(\displaystyle 3x=9\)

\(\displaystyle x=3\)

First we're going to get all the natural logs on one side of the equation:

\(\displaystyle ln(x-2) + ln(x) - ln(2x+9)= 0\)

Next, we're going to combine all the terms into one natural log:

\(\displaystyle ln(\frac{(x-2)(x)}{2x+9}) = 0\)

Now we can change to exponent form:

\(\displaystyle \frac{(x-2)(x)}{2x+9} = e^0\)

Anything raised to the \(\displaystyle 0\)th power equals \(\displaystyle 1\), which helps us simplify:

\(\displaystyle \frac{(x-2)(x)}{2x+9} = 1\)

\(\displaystyle (x-2)(x) = 2x+9\)

\(\displaystyle x^2-2x = 2x+9\)

\(\displaystyle x^2-9=0\)

From here, we can factor and solve:

\(\displaystyle (x+3)(x-3)=0\)

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

We have to notice, however, that \(\displaystyle -3\) isn't a valid answer because if we were to plug it into the original formula we would have a negative value in a logarithm. 

The first thing we can do is write the coefficients in front of the logs as exponents of the terms inside the logs:

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

Next, we combine the logs:

\(\displaystyle log(\frac{3x+1}{x^2})=1\)

Now we can change the equation into exponent form:

\(\displaystyle \frac{(3x+1)}{x^2}=10^1\)

When we simplify, we'll also move all the terms to one side of the equation:

\(\displaystyle 10x^2-3x-1=0\)

From here, we factor to get our solutions:

\(\displaystyle (5x+1)(2x-1)=0\)

\(\displaystyle x=\frac{1}{2}, -\frac{1}{5}\)

The last thing we have to do is check our answers.  \(\displaystyle \frac{1}{2}\) doesn't raise any problems, but \(\displaystyle -\frac{1}{5}\) does.  If we plug it back into the original equation, we would be evaluating a negative value in a log, which we can't do.

In order to solve for the logs, we will need to write the log properties as follows:

\(\displaystyle x^y = z\) and \(\displaystyle log_{x}z = y\)

This means that:

\(\displaystyle log_{2}(64) = 6\)

\(\displaystyle log_{4}(64) = 3\)

Replace the values into the expression.

\(\displaystyle 2log_{2}(64)-3log_{4}(64)= 2(6)-3(3) = 12-9 =3\)

The answer is:  \(\displaystyle 3\)

First, we rewrite the equation as:

\(\displaystyle log_e(e^{5x})=2\)

We can use log properties to simplify:

\(\displaystyle 5x=2\)

From here, simple algebra is used to solve:

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