The first thing we can do is combine all the log terms on the right side of the equation:

\(\displaystyle 2log(x)=log(\frac{9*8}{18})\)

Next, we can take the coefficient from the left term and make it an exponent:

\(\displaystyle log(x^2)=log(4)\)

Now we can cancel the logs from both sides:

\(\displaystyle x^2=4\)

\(\displaystyle x=\pm2\)

When we put \(\displaystyle 2\) back into the original question, we don't have problems.  When we try it with \(\displaystyle -2\) however, we get errors, so that's not a valid answer:

\(\displaystyle x=2\)

First, we take the coefficient, \(\displaystyle 3\), and make it an exponent:

\(\displaystyle log (x^3) = log (64)\)

Now we can cancel the logs:

\(\displaystyle x^3=64\)

\(\displaystyle x=\pm4\)

When we check our answers, however, we notice that \(\displaystyle -4\) results in errors, so that's not a valid answer:

\(\displaystyle x=4\)

First, we combine the log terms on the left of the equation:

\(\displaystyle log_3 ((x)(x-7)) = log_3 (8)\)

Now we can cancel the logs on each side:

\(\displaystyle x^2-7x=8\)

We can subtract \(\displaystyle 8\) from each side to set the equation equal to \(\displaystyle 0\).  this will give us a nice quadratic equation to solve:

\(\displaystyle x^2-7x-8=0\)

\(\displaystyle (x-8)(x+1)=0\)

\(\displaystyle x=8, -1\)

Notice that \(\displaystyle -1\) is not a valid answer, because if we plug it into the original equation then we would be taking the log of a negative number, which we can't do.  Our only solution is:

\(\displaystyle x=8\)

The first thing we can do is move both log functions to one side of the equation:

\(\displaystyle log_2(x-4) + log_2 (x+3)= 3\)

Then we can combine the log functions (remember, when you add logs, we multiply the terms inside):

\(\displaystyle log_2((x-4)(x+3))= 3\)

Now we can rewrite the equation in exponent form (and FOIL the multiplied terms):

\(\displaystyle x^2-x-12=2^3\)

We can collect all the terms on one side of the equation, and then solve the quadratic:

\(\displaystyle x^2-x-20=0\)

\(\displaystyle (x-5)(x+4)=0\)

\(\displaystyle x=-4, 5\)

However, if we plug \(\displaystyle -4\) into the initial equation, we would be taking the log of a negative number, which we can't do, so it's not a valid solution:

\(\displaystyle x=5\)

We first put both logs on one side of the equation:

\(\displaystyle log_7(6x-16)-log_7(x-1)=0\)

Now we combine the log terms (remember, when we subtract logs we divide the terms inside):

\(\displaystyle log_7(\frac{(6x-16)}{(x-1)})=0\)

We can now rewrite the equation in exponential form:

\(\displaystyle \frac{6x-16}{x-1}=7^0\)

Anything raised to the \(\displaystyle 0\) power is \(\displaystyle 1\), and now we can solve algebraically:

\(\displaystyle 6x-16=x-1\)

\(\displaystyle 5x=15\)

\(\displaystyle x=3\)

We start by rewriting the equation in exponential form:

\(\displaystyle 3x-8=5^2\)

Now we can simplify:

\(\displaystyle 3x-8=25\)

\(\displaystyle 3x=33\)

\(\displaystyle x=11\)

We can either do this the long and proper way, or the simple and easy way.

The long way:

First, we move both logs to the same side of the equation:

\(\displaystyle log (2x)- log (24) =0\)

Now we can combine the logs (reminder, when you subtract logs, you divide the terms inside of them):

\(\displaystyle log(\frac{(2x)}{24})=0\)

Let's rewrite the equation in exponential form:

\(\displaystyle \frac{2x}{24}=10^0\)

Anything raised to the \(\displaystyle 0\) power equals \(\displaystyle 1\), so we can simplify and solve from here:

\(\displaystyle \frac{2x}{24}=1\)

\(\displaystyle 2x=24\)

\(\displaystyle x=12\)

The short way:

First, we cancel the log terms (because the base is the same, and all we have are the log terms):

\(\displaystyle 2x=24\)

Then we divide by \(\displaystyle 2\):

\(\displaystyle x=12\)

\(\displaystyle \log_{b}a\) is equal to \(\displaystyle b^{x}=a\),

so in this case it is 

\(\displaystyle 8^{x}=64\),

and 

\(\displaystyle x=2\)