First, we can combine the log terms:

Now we can change to exponent form (remember, if a log doesn't specifically have a base, then it's base 10):

We need to set the equation equal to  in order to solve the quadratic equation, so we combine the  terms and subtract :

Then we factor and solve for :

Lastly, we have to check our answers.  When we plug  in to the original equation, everything comes out well.  However, when we use  we get errors (because you can't take the log of a negative number). Therefore, we only have 1 solution: .

First, we combine the logs:

Next, we can rewrite the equation in exponent form:

Now we use algebra to set the equation equal to :

We then factor and solve for :

Lastly, we have to check our answers.  Using  in the original equation results in errors (because you can't take the log of a negative number).   works out fine though, so we have 1 solution.

First, we add  to each side of the equation:

Next, we divide each side by :

Now, we write the equation in exponent form and solve for :

Plugging our answer doesn't produce any errors, so it's a valid answer!

First, we subtract  from each side:

Then we divide each side by :

Now we write the equation in exponent form (remember, "ln" has a base of ):

We could solve numerically, but the answer would be messy, so let's leave it like this.  If we plug it back into the original equation, we see that there are no errors.

First, combine the ln terms on the left of the equation:

Next, we can make each term an exponent of :

Remember, a log that's an exponent of it's base cancels out, so our equation becomes:

Now we FOIL the terms on the left, and set the equation equal to .

Lastly, we factor (or use the quadratic equation) to find our solutions:

If we check our solutions, we notice that  produces an error (you can't take the log of a negative number), so  is our only solution.

The first thing we can do is use the change of base formula on the term on the left so that all our logs are in base :

, so we can simplify that, and then multiply the denominator to the other side:

We can subtract  from each side and simplify a bit:

It can be tough to see what to do here, but making a substitution can clear it up a bit:

Now we can see that the form of the equation is a quadratic.  We can factor and solve:

We now plug our answers back into our substitution.  Starting with :

Write it in exponent form and solve:

Plugging this back into the original equation, we don't get any errors, so it's a valid solution.  Now we do the same with :

We could solve this numerically, but the answer wouldn't be tidy, so let's leave it as is.  Plugging it back into the original equation, we don't get any errors, so it's also a valid solution.

The first thing we can do is get rid of the exponent in the second log, and then divide both sides by that coefficient:

It can be a pain to work with logs of different bases, so let's change the base  log to base :

We can now take the denominator, expand, and simplify:

Now, we combine the log terms:

Taking the square root of each side:

We have to solve for both the plus, and the minus.  Let's start with the negative:

And now the positive:

Plugging both of these answers into the original equation produces no errors, so both answers are valid.

First, we can use the log product rule to combine the log terms:

Now we write the equation in exponential form:

If we plug  into the original equation, we get errors (we can't take the log of a negative value).   is fine though, and is our only solution.

We can combine the logs on the right side of the equation while also using the product rule on the log on the left side:

Since the logs have the same base, we can equate the algebraic equations equal to each other:

Now we can solve the quadratic equation:

Plugging this into the original equation doesn't produce errors, so it's our single solution.

First we take the natural log of both sides of the equation:

Remember, taking the natural log of  with an exponent equals that exponent, and the natural log of  is :

Form here, we solve algebraically: