Applying Combined Exponents Rules: Multiplying Exponents with the same base

$7^5\cdot7^{-6}=\text{?}$

We begin by using the rule for multiplying exponents. (the multiplication between terms with identical bases):

$a^m\cdot a^n=a^{m+n}$We then apply it to the problem:

$7^5\cdot7^{-6}=7^{5+(-6)}=7^{5-6}=7^{-1}$When in a first stage we begin by applying the aforementioned rule and then continue on to simplify the expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression obtained in the previous step:

$7^{-1}=\frac{1}{7^1}=\frac{1}{7}$We then summarise the solution to the problem: $7^5\cdot7^{-6}=7^{-1}=\frac{1}{7}$Therefore, the correct answer is option B.

$$$\frac{1}{7}$

$12^4\cdot12^{-6}=\text{?}$

We begin by using the power rule of exponents; for the multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it to the given problem:

$12^4\cdot12^{-6}=12^{4+(-6)}=12^{4-6}=12^{-2}$When in a first stage we apply the aforementioned rule and then simplify the subsequent expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression that we obtained in the previous step:

$12^{-2}=\frac{1}{12^2}=\frac{1}{144}$Lastly we summarise the solution to the problem: $12^4\cdot12^{-6}=12^{-2} =\frac{1}{144}$Therefore, the correct answer is option A.

$\frac{1}{144}$

Solve the exercise:

$Y^2+Y^6-Y^5\cdot Y=$

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it in the problem:

$Y^2+Y^6-Y^5\cdot Y=Y^2+Y^6-Y^{5+1}=Y^2+Y^6-Y^6=Y^2$When we apply the previous property to the third expression from the left in the sum, and then simplify the total expression by adding like terms.

Therefore, the correct answer is option D.

$Y^2$

$(-\frac{1}{8})^8\cdot(-\frac{1}{8})^{-3}=?$

$-8^{-5}$

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=\text{?}$

$\frac{1}{9^{-3}}$

$\frac{1}{-3}\cdot3^{-4}\cdot5^3=\text{?}$

$-\frac{5^3}{3^5}$

$4^{2x}\cdot\frac{1}{4}\cdot4^{-2}=\text{?}$

$\frac{1}{4^{3-2x}}$