$\displaystyle 9 = 3^{2}$ and $\displaystyle 27 = 3 ^{3}$, so, 

$\displaystyle 9 ^{x+6} = 27 ^{2x}$

can be rewritten as

$\displaystyle ( 3 ^{2}) ^{x+6} = (3 ^{3})^{2x}$

Applying the Power of a Power Rule,

$\displaystyle 3 ^{2 \cdot (x+6)} = 3 ^{3 \cdot 2x}$

$\displaystyle 3 ^{2 x+12} = 3 ^{6x}$

$\displaystyle 2x+12 = 6x$

$\displaystyle 2x+12 - 2x = 6x - 2x$

$\displaystyle 12 = 4x$

$\displaystyle \frac{12 }{4}= \frac{4x}{4}$

$\displaystyle x = 3$

Rewrite the base of the right side.

$\displaystyle 9=3^2$

$\displaystyle 3^9 = 3^{2(3x-3)}$

$\displaystyle 9=2(3x-3)$

Simplify the right side.

$\displaystyle 9= 6x-6$

Add 6 on both sides.

$\displaystyle 9+6= 6x-6+6$

$\displaystyle 15 = 6x$

Divide by 6 on both sides.

$\displaystyle \frac{15}{6} =\frac{ 6x}{6}$

The answer is:  $\displaystyle \frac{5}{2}$

Change the base of the left side to base two.

$\displaystyle (2^3)^3 = 2^{2x}$

The equation becomes:

$\displaystyle 2^9 = 2^{2x}$

Set the exponents equal since they have similar bases.

$\displaystyle 9=2x$

Divide by 2 on both sides.

$\displaystyle \frac{9}{2}=\frac{2x}{2}$

The answer is:  $\displaystyle \frac{9}{2}$

Start by distributing the exponent in both the numerator and the denominator. Recall that when an exponent is raised to the exponent, you will need to multiply the two numbers together.

$\displaystyle \frac{(a^3b^2c^7)^{-3}}{(a^7b^4c^{-2})^2}=\frac{a^{-9}b^{-6}c^{-21}}{a^{14}b^8c^{-4}}$

Next, recall that when you have numbers with different exponents, but the same base, subtract the exponent found in the denominator from the exponent in the numerator.

$\displaystyle \frac{a^{-9}b^{-6}c^{-21}}{a^{14}b^8c^{-4}}=a^{-23}b^{-14}c^{-17}$

Recall that you can flip the fraction to make the exponents positive.

$\displaystyle a^{-23}b^{-14}c^{-17}=\frac{1}{a^{23}b^{14}c^{17}}$

The first thing we need to do is find a common base.  This can be tricky to do, but guessing and checking a little shows that:

$\displaystyle (\frac{1}{216})=(\frac{1}{36})^{\frac{3}{2}}$

Plugging that back in to the original equation:

$\displaystyle (\frac{1}{36})^{x+3}=(\frac{1}{36})^{\frac{3}{2}}$

Now that our bases are the same, we can cancel them:

$\displaystyle x+3=\frac{3}{2}$

From here, it's much easier to solve using simple algebra:

$\displaystyle x=-\frac{3}{2}$

First, we gather all the constants on one side of the equation:

$\displaystyle log_2 (5x) = 4$

Next, we rewrite the equation in exponential form:

$\displaystyle 2^4=5x$

Now we can simplify the exponent:

$\displaystyle 16=5x$

And finally, divide:

$\displaystyle \frac{16}{5}=x$

Start by combining log terms.  Remember, if you subtract logs, you divide the terms inside them:

$\displaystyle log_9(\frac{10}{5x})=1$

Now we can rewrite the equation in exponential form:

$\displaystyle 9^1=\frac{10}{5x}$

Finally, we need to get the variable in the numerator, and then alone:

$\displaystyle 45x=10$

$\displaystyle x=\frac{10}{45}=\frac{2}{9}$

First, we combine log terms by subtracting them.  Remember, when you subtract logs, you divide the terms inside them:

$\displaystyle log_3(\frac{4x-10}{8})=log_340$

Now, because the bases of the logs match on either side of the equation, we can cancel them out:

$\displaystyle \frac{4x-10}{8}=40$

From here, we use simple algebra to solve:

$\displaystyle 4x-10=320$

$\displaystyle 4x=330$

$\displaystyle x=\frac{165}{2}$

First, subtract the natural log terms:

$\displaystyle ln(\frac{x}{4})=5$

Now rewrite the equation in exponential form:

$\displaystyle e^5=\frac{x}{4}$

Finally, isolate the variable:

$\displaystyle 4e^5=x$

We can start by canceling the logs, because they both have the same base:

$\displaystyle 2-x=-2x-2$

Now we can collect constants on one side of the equation, and variables on the other:

$\displaystyle x=-4$