We can rewrite the right side by changing the base.

$\displaystyle \frac{1}{8} = (\frac{1}{2})^3$

Rewrite the right side using the new base.

$\displaystyle (\frac{1}{2})^{5x+2} =[(\frac{1}{2})^3]^{3x}$

Now that the bases are similar, we can set the exponents equal to each other.

$\displaystyle 5x+2 = 9x$

Subtract $\displaystyle 5x$ on both sides.

$\displaystyle 5x+2 -5x= 9x-5x$

$\displaystyle 2=4x$

Divide by four on both sides.

$\displaystyle \frac{2}{4}=\frac{4x}{4}$

Reduce the fractions.

The answer is:  $\displaystyle \frac{1}{2}$

We will need to convert the bases on both sides to base three.

$\displaystyle \frac{1}{3} = 3^{-1}$

$\displaystyle \frac{1}{9}=(\frac{1}{3})^2$

Rewrite the equation.

$\displaystyle 3[(3)^{-1}]^{3x+1}= (3^{-2})^{10}$

Simplify both sides.

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

On the left side, notice that a lone three is the coefficient of the base.  We will need to write the left side such that:

$\displaystyle 3(3)^{-1(3x+1)} = 3^1\cdot (3)^{-1(3x+1)} = 3^{1+(-1)(3x+1)}$

Recall that if the powers of the same base are multiplied, the powers can be added.

Simplify the exponent.

$\displaystyle 3^{1+(-1)(3x+1)} = 3^{1-3x-1} = 3^{-3x}$

The equation becomes:  $\displaystyle 3^{-3x} = 3^{-20}$

Set the exponential terms equal to each other now that the bases are equal.

$\displaystyle -3x=-20$

Divide by negative three on both sides.

$\displaystyle \frac{-3x}{-3}=\frac{-20}{-3}$

The answer is:  $\displaystyle \frac{20}{3}$

Change the base of the second term to base two.

$\displaystyle 8=2^3$

$\displaystyle 2^{3x+3} =(2^3)^{-5x}$

Simplify the right side by multiplying the exponents.

$\displaystyle 2^{3x+3} =(2)^{-15x}$

Now that the bases are common, the exponents can be set equal to each other.

$\displaystyle 3x+3 = -15x$

Subtract $\displaystyle 3x$ from both sides.

$\displaystyle 3x+3-3x = -15x-3x$

$\displaystyle 3=-18x$

Divide by negative 18 on both sides.

$\displaystyle \frac{3}{-18}=\frac{-18x}{-18}$

The answer is:  $\displaystyle -\frac{1}{6}$

To evaluate this equation, we will need to change the base of the left side.

$\displaystyle 4=2^2$

Rewrite the equation.

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

$\displaystyle 2^{12}= 2^{3x-3}$

Now that both bases are similar, we can set the exponents equal to each other.

$\displaystyle 12=3x-3$

Add 3 on both sides.

$\displaystyle 12+3=3x-3+3$

$\displaystyle 15=3x$

Divide by three on both sides.

The answer is:  $\displaystyle 5$

Rewrite both sides of the equation so that we have same bases.

$\displaystyle 2(\frac{1}{2})^{7x-1}= (\frac{1}{4})^{3x+2}$

$\displaystyle 2=2^1$

$\displaystyle \frac{1}{2} = 2^{-1}$

$\displaystyle \frac{1}{4} = 2^{-2}$

$\displaystyle 2^1\cdot (2^{-1})^{7x-1}= (2^{-2})^{3x+2}$

Simplify the exponents.

$\displaystyle 2^1\cdot (2^{-7x+1})= 2^{-6x-4}$

Add the exponents on the left side.

$\displaystyle 2^{-7x+2}= 2^{-6x-4}$

Now that both sides have same bases, we can set the exponential terms equal.

$\displaystyle -7x+2 = -6x-4$

Add $\displaystyle 7x$ on both sides.

$\displaystyle -7x+2+7x = -6x-4+7x$

$\displaystyle 2 = x-4$

Add four on both sides.

$\displaystyle 2 +4= x-4+4$

The answer is:  $\displaystyle 6$

In order to solve this, we will need to rewrite the fractional base to base three.

$\displaystyle \frac{1}{27} = 3^{-3}$

Rewrite the equation using this base.

$\displaystyle 3^4 = (3^{-3})^{9x-3}$

Now that both bases are common, we can set up an equation where the powers are equal.

$\displaystyle 4=-3(9x-3)$

Use distribution to simplify the right side.

$\displaystyle 4=-27x+9$

Subtract nine from both sides.

$\displaystyle 4-9=-27x+9-9$

$\displaystyle -5 = -27x$

Divide by negative 27 on both sides.

$\displaystyle \frac{-5 }{-27}= \frac{-27x}{-27}$

The answer is:  $\displaystyle \frac{5}{27}$

To solve this exponential equation, we will need to change the base of the second term.  Notice that both the numerator and denominator are the values of the left fraction cubed.  

$\displaystyle \frac{8}{27} = (\frac{2}{3})^3$

Rewrite the equation.

$\displaystyle (\frac{2}{3})^{3x-8} =( \frac{2}{3})^{3(-x-2)}$

With common bases, we can set the exponents equal to each other.

$\displaystyle 3x-8 = 3(-x-2)$

Distribute the right side.

$\displaystyle 3x-8 = -3x-6$

Add $\displaystyle 3x$ on both sides.

$\displaystyle 3x-8 +(3x)= -3x-6+(3x)$

$\displaystyle 6x-8 = -6$

Add 8 on both sides.

$\displaystyle 6x-8 +8= -6+8$

$\displaystyle 6x=2$

Divide by six on both sides.

$\displaystyle \frac{6x}{6}=\frac{2}{6}$

The answer is:  $\displaystyle \frac{1}{3}$

In order to solve this equation, we will need to change the base of the left side.

Rewrite one-ninth as base three.

$\displaystyle \frac{1}{9} = 3^{-2}$

Rewrite the equation.

$\displaystyle (3^{-2}) ^{81} = 3^{2x-10}$

Now that both bases are common, we can set both powers equal.

$\displaystyle (-2)(81) = 2x-10$

Simplify the left side and solve for x.

$\displaystyle -162 = 2x-10$

Add 10 on both sides, and then divide by two.

$\displaystyle -162 +10= 2x-10+10$

$\displaystyle \frac{-152}{2}=\frac{2x}{2}$

The answer is:  $\displaystyle -76$

Change the base of the right side.

$\displaystyle \frac{1}{2} =2^{-1}$

Rewrite the base of the right side.

$\displaystyle 2^{5x+2} = (2^{-1})^{10x+3}$

Set the exponents equal to each other now that both of the bases are alike.

$\displaystyle 5x+2 = -1(10x+3)$

Simplify both sides.

$\displaystyle 5x+2 = -10x-3$

Add $\displaystyle 10x$ on both sides.

$\displaystyle 5x+2 +10x= -10x-3+10x$

$\displaystyle 15x+2= -3$

Subtract two on both sides.

$\displaystyle 15x+2-2= -3-2$

$\displaystyle 15x=-5$

Divide by 15 on both sides.

$\displaystyle \frac{15x}{15}=\frac{-5}{15}$

Reduce both fractions.

The answer is:  $\displaystyle -\frac{1}{3}$

The bases on both sides of the equation are equal.  This means we can set the powers equal to each other.

$\displaystyle 10x= -9x-19$

Add $\displaystyle 9x$ on both sides.

$\displaystyle 10x+9x= -9x-19+9x$

$\displaystyle 19x = -19$

Divide by 19 on both sides.

$\displaystyle \frac{19x}{19} = \frac{-19}{19}$

The answer is:  $\displaystyle -1$