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:

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

First we divide both sides of the equation by \(\displaystyle 2\):

\(\displaystyle log_x(64)=\frac{3}{2}\)

Next, we write the equation in exponential form:

\(\displaystyle x^{\frac{3}{2}}=64\)

Then we solve for \(\displaystyle x\) by taking the cube root, and then squaring \(\displaystyle 64\):

\(\displaystyle x=16\)

First we can divide each side by \(\displaystyle 2\):

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

Next, we rewrite in exponent form:

\(\displaystyle (x+1)^2=9\)

From here, we simply take the square root of \(\displaystyle 9\), and then subtract \(\displaystyle 1\):

\(\displaystyle x=2\)

First we rewrite the equation in exponential form:

\(\displaystyle x^4=16\)

Now we take the cube root of \(\displaystyle 16\):

\(\displaystyle x=2\)

First we subtract \(\displaystyle 3\) from both sides:

\(\displaystyle 3log(2x)=9\)

Then we divide both sides by \(\displaystyle 3\):

\(\displaystyle log(2x)=3\)

Now it would help if we wrote the equation in exponential form (remember, if the log doesn't show a base, it's base 10):

\(\displaystyle 2x=10^3\)

Finally, we use algebra to solve:

\(\displaystyle 2x=1000\)

\(\displaystyle x=500\)