a⁴ = a² * a²  and  b⁶= b² * b²  * b²

Therefore a⁴ + b⁶ = 35 * 35 + 52 * 52 * 52 = 1,225 + 140,608 = 141,833

Since we have two \(\displaystyle x\)’s in \(\displaystyle 9^{(x + 5)} + 3^{2(x + 5)}\) we will need to combine the two terms.

For \(\displaystyle 3^{2(x + 5) }\) this can be rewritten as

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

So we have \(\displaystyle 9^{ (x + 5) }+ 9^{ (x + 5)} = 162\).

Or \(\displaystyle 2 (9^{ (x + 5)}) = 162\)

Divide this by \(\displaystyle 2\): \(\displaystyle 9^{ (x + 5)} = 81 = 9^ 2\)

Thus \(\displaystyle x +5 = 2\) or \(\displaystyle x = -3\)

*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.

The answer is 5. 

8 + 2^x+1 = 72

      2^x+1 = 64

      2^x+1 = 2⁶

      x + 1 = 6

           x = 5

We want to simplify 5^b – 5^(b–1) – 5^(b–1) – 5^(b–1) – 5^(b–1) – 5^(b–1) .

Notice that we can collect the –5^(b–1) terms, because they are like terms. There are 5 of them, so that means we can write –5^(b–1) – 5^(b–1) – 5^(b–1) – 5^(b–1) – 5^(b–1) as (–5^(b–1))5.

To summarize thus far:

5^b – 5^(b–1) – 5^(b–1) – 5^(b–1) – 5^(b–1) – 5^(b–1) = 5^b +(–5^(b–1))5

It's important to interpret –5^(b–1) as (–1)5^(b–1) because the –1 is not raised to the (b – 1) power along with the five. This means we can rewrite the expression as follows:

5^b +(–5^(b–1))5 = 5^b + (–1)(5^(b–1))(5) = 5^b – (5^(b–1))(5)

Notice that 5^(b–1) and 5 both have a base of 5. This means we can apply the property of exponents which states that, in general, a^ba^c = a^b+c. We can rewrite 5 as 5¹ and then apply this rule.

5^b – (5^(b–1))(5) = 5^b – (5^(b–1))(5¹) = 5^b – 5^(b–1+1)

Now, we will simplify the exponent b – 1 + 1 and write it as simply b.

5^b – 5^(b–1+1) = 5^b – 5^b = 0

The answer is 0.

\(\displaystyle x^4=(x^2)(x^2)=11\cdot 11=121\)

Step 1: \(\displaystyle \left ( x^{-2}x^{5} \right )= x^{3}\)

Step 2: \(\displaystyle \left ( y^{3}y^{-4} \right )= y^{-1}= \frac{1}{y}\)

Step 3: (Correct Answer): \(\displaystyle \frac{x^{3}}{y}\)

Step 1: \(\displaystyle \frac{y^{5}}{\left ( x^{3}x^{2} \right )\left \right )y^{-1}}\)

Step 2: \(\displaystyle \frac{\left ( y^{5}y^{1} \right )}{x^{3}x^{2}}\)

Step 3:\(\displaystyle \frac{y^{6}}{x^{5}}\)

Step 1:\(\displaystyle \frac{1}{\left ( -11 \right )^{12}\left ( -11 \right )^{8}}\)

Step 2: The above is equal to \(\displaystyle \frac{1}{\left ( -11 \right )^{20}}\)

\(\displaystyle -\left ( -3 \right )^{0}= -1\)

\(\displaystyle (-3^{0})=-1\)

 \(\displaystyle -1-(-1)=-1+1=0\)

\(\displaystyle 3\sqrt{18} = 3\sqrt{2\star 3^{2}} = 9\sqrt{2}\)

Similarly \(\displaystyle 2\sqrt{50} = 2\sqrt{2\star 5^{2}} = 10\sqrt{2}\)

So \(\displaystyle 9\sqrt{2} + 10\sqrt{2}= 19\sqrt{2}\)