Solve for x:

\(\displaystyle 3^{3x+1}=81\)

Step 1: Represent \(\displaystyle 81\) exponentially with a base of \(\displaystyle 3\)

\(\displaystyle 3\cdot 3\cdot 3\cdot 3=81\), therefore \(\displaystyle 3^4=81\)

Step 2: Set the exponents equal to each other and solve for x

\(\displaystyle 3^{3x+1}=3^4\)

\(\displaystyle 3x+1 = 4\)

\(\displaystyle 3x = 3\)

\(\displaystyle {\color{Red} x=1}\)

Rewrite \(\displaystyle 256\) in exponential form with a base of \(\displaystyle 4\):

\(\displaystyle 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=256\)

\(\displaystyle 2^8=256\)

Solve for \(\displaystyle x\) by equating exponents:

\(\displaystyle 2^{3x+2}=2^{8}\)

\(\displaystyle 3x+2=8\)

\(\displaystyle 3x=6\)

\(\displaystyle {\color{Red} x=2}\)

Represent \(\displaystyle 1024\) in exponential form using a base of \(\displaystyle 4\):

\(\displaystyle 4\cdot 4\cdot 4\cdot 4\cdot 4=1024\)

\(\displaystyle 4^5=1024\)

Solve for \(\displaystyle x\) by equating exponents:

\(\displaystyle 4^{2x+1}=4^5\)

\(\displaystyle 2x+1=5\)

\(\displaystyle 2x=4\)

\(\displaystyle {\color{Red} x=2}\)

The smallest perfect square between 100 and 1,000 inclusive is 100 itself, since \(\displaystyle 10^{2} = 100\). The largest can be found by noting that \(\displaystyle \sqrt{1,000} \approx 31.6\); this makes \(\displaystyle 31^{2} = 961\) the greatest perfect square in this range. 

Since the squares of the integers from 10 to 31 all fall in this range, this makes \(\displaystyle 31 - 10 + 1 = 22\) perfect squares. 

To solve this expression, remove the outer exponent and expand the terms.

\(\displaystyle (x^2y^3)^3=x^2\cdot x^2 \cdot x^2 \cdot y^3 \cdot y^3 \cdot y^3\)

By exponential rules, add all the powers when multiplying like terms.

The answer is:  \(\displaystyle x^6y^9\)

The first step in solving for x is to simplify the right side:

\(\displaystyle 2^22^x=2^{2+x}\).

Next, we can re-express the left side as an exponential with 2 as the base.

\(\displaystyle 16=2^4\)

Now set the new left side equal to the new right side.

\(\displaystyle 2^4=2^{2+x}\)

With the bases now being the same, we can simply set the exponents equal to each other.

\(\displaystyle 4=2+x\)

\(\displaystyle x=2\)

To solve for x, we need to simplify both sides in order to make the equation simpler to solve.

\(\displaystyle 64\) can be rewritten as \(\displaystyle 4^3\), and \(\displaystyle \frac{4^6}{4^x}\) can be written as \(\displaystyle 4^{6-x}\).

Setting the two sides equal to each other gives us

\(\displaystyle 4^3=4^{6-x}\)

Since the bases are the same we can set the exponents equal to each other.

\(\displaystyle 3=6-x\)

\(\displaystyle x=3\)

\(\displaystyle 2^5\) is expanded to \(\displaystyle 2*2*2*2*2\).

We will simply multiply the values in order to get the answer:

\(\displaystyle 32\)

When we expand exponents, we simply repeat the base by the exponential value.

Therefore:

\(\displaystyle 3^4=3*3*3*3\)

When we expand exponents, we simply repeat the base by the exponential value.

Therefore:

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