To solve this equation, we will need to convert the 100 into base ten.

\(\displaystyle 100=10^2\)

Rewrite the number using this base.

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

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

\(\displaystyle 3x+5 =2(2x)\)

Simplify this equation.

\(\displaystyle 3x+5=4x\)

Subtract \(\displaystyle 3x\) on both sides.

\(\displaystyle 5=x\)

The answer is:  \(\displaystyle 5\)

To be able to set the powers equal to each other, we will need common bases.

Convert eight into two cubed.

\(\displaystyle 2^{3(x-1)} = 2^{3(6x)}\)

Set the powers equal to each other.

\(\displaystyle 3(x-1)= 3(6x)\)

Divide by three on both sides.

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

Subtract \(\displaystyle x\) from both sides.

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

\(\displaystyle -1= 5x\)

Divide by five on both sides.

\(\displaystyle \frac{-1}{5}= \frac{5x}{5}\)

The answer is:  \(\displaystyle -\frac{1}{5}\)

To solve this equation, I would first rewrite 8 as a base of 2:

\(\displaystyle 8^3=(2^3)^3\)

Now, plug back into the equation and simplify. When there are two exponents next to each other like this, multiply them:

\(\displaystyle 2^{x}=2^9\)

Since the bases are the same, you can set the exponents equal to each other:

\(\displaystyle x=9\)

In order to solve this, the bases of the powers will need to be converted.  Notice that both terms can be rewritten as base three.

\(\displaystyle \frac{1}{3}= 3^{-1}\)

\(\displaystyle 27= 3^3\)

Rewrite the equation.

\(\displaystyle (3)^{-1(3x-1)} = 3^{3(5x)}\)

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

\(\displaystyle -1(3x-1)= 3(5x)\)

Divide negative one on both sides.  This will move the negative to the other side.

\(\displaystyle 3x-1 =-15x\)

Subtract \(\displaystyle 3x\) on both sides.

\(\displaystyle 3x-1-3x =-15x-3x\)

\(\displaystyle -1=-18x\)

Divide by negative 18 on both sides.

\(\displaystyle \frac{-1}{-18}=\frac{-18x}{-18}\)

The answer is:  \(\displaystyle \frac{1}{18}\)

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

\(\displaystyle 16=2^4\)

Rewrite the equation.

\(\displaystyle 2^{100} = 2^{4(50x)}\)

With similar bases, the exponential powers can be set equal to each other.

\(\displaystyle 100= 200x\)

Divide by 200 on both sides.

\(\displaystyle \frac{100}{200}= \frac{200x}{200}\)

Simplify both sides.

The answer is:  \(\displaystyle \frac{1}{2}\)

In order to solve this equation, we will need similar bases to continue.  Notice that both bases have a common base of two.  We can rewrite each base using two to the power of a certain number to express the base.

\(\displaystyle \frac{1}{2} = 2^{-1}\)

\(\displaystyle 32= 2^5\)

Rewrite the equation.

\(\displaystyle (2)^{-1(3x+2)}= 2^{5(5x-7)}\)

With bases similar, we can set the powers equal to each other.

\(\displaystyle -1(3x+2) = 5(5x-7)\)

\(\displaystyle -3x-2 = 25x-35\)

Add \(\displaystyle 3x\) on both sides.

\(\displaystyle -3x-2 +3x= 25x-35+3x\)

\(\displaystyle -2=28x-35\)

Add 35 on both sides.

\(\displaystyle -2+35=28x-35+35\)

\(\displaystyle 33=28x\)

Divide by 28 on both sides.

\(\displaystyle \frac{33}{28}=\frac{28x}{28}\)

The answer is:  \(\displaystyle \frac{33}{28}\)

In order to solve this equation, we will need to convert the bases to a common base.

The one-fifth and 125 can be rewritten as certain powers of five.  Rewrite the numbers.

\(\displaystyle \frac{1}{5} = 5^{-1}\)

\(\displaystyle 125 = 5^3\)

Replace the numbers with common bases.

\(\displaystyle (5^{-1})^{3x-2} =(5^3)^{9}\)

Now that the bases are common, we can set the powers equal to each other.

\(\displaystyle -1(3x-2) = 27\)

Divide by negative one on both sides.  The equation becomes:

\(\displaystyle 3x-2 =-27\)

Add two on both sides.

\(\displaystyle 3x-2+2 =-27+2\)

\(\displaystyle 3x=-25\)

Divide by three on both sides.

\(\displaystyle \frac{3x}{3}=\frac{-25}{3}\)

The answer is:  \(\displaystyle -\frac{25}{3}\)

The bases of \(\displaystyle e\) are already equal.  There is no need to rewrite the right side of the equation using a fraction.

Since the bases are equal, we can set the powers equal to each other.

\(\displaystyle 2x+3=-2x\)

Add \(\displaystyle 2x\) on both sides.

\(\displaystyle 2x+3+2x=-2x+2x\)

\(\displaystyle 4x+3=0\)

Subtract three on both sides.

\(\displaystyle 4x+3-3=0-3\)

\(\displaystyle 4x=-3\)

Divide by four on both sides.

\(\displaystyle \frac{4x}{4}=\frac{-3}{4}\)

The answer is:  \(\displaystyle -\frac{3}{4}\)

Notice that we can rewrite the right side of the equation as an exponent.

The equation \(\displaystyle e^{6x-4} = \frac{1}{e}\) becomes:

\(\displaystyle e^{6x-4} = e^{-1}\)

Now that the bases of the exponential term are equal, the powers can be set equal to each other.

\(\displaystyle 6x-4 = -1\)

Add 4 on both sides.

\(\displaystyle 6x-4 +4= -1+4\)

Simplify the equation.

\(\displaystyle 6x=3\)

Divide by six on both sides.

\(\displaystyle \frac{ 6x}{6}=\frac{3}{6}\)

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

The answer is:  \(\displaystyle \frac{1}{2}\)

Change the base on the right side to base ten.

\(\displaystyle 100= 10^2\)

Replace the hundred with this term.

\(\displaystyle 10^{2x-3}= 10^{2(9-3x)}\)

Set the powers equal to each other now that the bases are equivalent.

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

Solve for the x-variable.  Distribute the two on the right side.

\(\displaystyle 2x-3 = 18-6x\)

Add \(\displaystyle 6x\) on both sides.

\(\displaystyle 2x-3+6x = 18-6x+6x\)

\(\displaystyle 8x-3 = 18\)

Add three on both sides.

\(\displaystyle 8x-3+3 = 18+3\)

\(\displaystyle 8x=21\)

Divide by eight on both sides.

\(\displaystyle \frac{8x}{8}=\frac{21}{8}\)

The answer is:  \(\displaystyle \frac{21}{8}\)