How to Solve Exponent Problems?

Here we have a negative exponent, so we have to start by converting that exponent to positive. Thus, applying the negative exponents' rule, we have to take the reciprocal of the base and change the exponent from negative to positive:

$latex 5^{-2}=\frac{1}{5^2}$

Now, we apply the exponent to 5 to simplify:

$latex \frac{1}{{{5}^2}}=\frac{1}{25}$

In this case, we have to apply the power of a power rule, where we multiply the exponents:

$latex {{({{3}^2})}^{-2}}={{3}^{-4}}$

Similar to the previous exercise, we apply the negative exponents' rule to change the exponent from negative 4 to positive 4:

$latex {{3}^{-4}}=\frac{1}{{{3}^4}}$

Applying the exponent, we have:

$latex \frac{1}{{{3}^4}}=\frac{1}{81}$

We have a division of powers with the same base. Therefore, we use the quotient rule. This rule tells us that we subtract the exponent of the denominator from the exponent of the numerator:

$latex \frac{{{4}^5}}{{{4}^3}}={{4}^{5-3}}$

$latex ={{4}^2}$

Applying the exponent we have:

$latex {{4}^2}=16$

In the numerator we have a multiplication of powers, but we cannot simplify because the base is not the same. However, we can rewrite the denominator by recognizing that 12 equals 4 × 3:

$latex \frac{{{4}^5}\times {{3}^5}}{{{12}^3}}=\frac{{{4}^5}\times {{3}^5}}{{{(4\times3)}^3}}$

$latex =\frac{{{4}^5}\times {{3}^5}}{{{4}^3}\times {{3}^3}}$

Now, we can apply the quotient rule to both base 3 powers and base 4 powers:

$latex \frac{{{4}^5}\times {{3}^5}}{{{4}^3}\times {{3}^3}}={{4}^{5-3}}\times {{3}^{5-3}}$

$latex ={{4}^2}\times {{3}^2}$

$latex =16\times 9$

$latex =144$

First, we can start by changing the negative exponent to positive by flipping the fraction over:

$latex {{\left(\frac{4}{{{5}^2}} \right)}^{-2}} \times \left(\frac{{{4}^2}}{{{5}^3}} \right)={{\left(\frac{{{5}^2}}{4} \right)}^2} \times \left(\frac{{{4}^2}}{{{5}^3}} \right)$

Now, we can apply the power of a power rule:

$latex {{\left(\frac{{{5}^2}}{4} \right)}^2} \times \left(\frac{{{4}^2}}{{{5}^3}} \right)=\frac{{{5}^4}}{{{4}^2}}  \times \frac{{{4}^2}}{{{5}^3}} $

We can rewrite as follows and apply the quotient rule:

$latex \frac{{{5}^4}}{{{4}^2}}  \times \frac{{{4}^2}}{{{5}^3}}=\frac{{{5}^4}}{{{5}^3}}  \times \frac{{{4}^2}}{{{4}^2}}$

$latex ={{5}^{4-3}}\times{{4}^{2-2}}$

Simplifying and applying the zero rule of exponents, we have:

$latex {{5}^{4-3}}\times{{4}^{2-2}}={{5}^1}\times {{4}^0}$

$latex =5\times 1$

$latex =5$

In this case, we have the variables x and y, but the rules of exponents are applied in the same way. Therefore, we start with the negative exponents' rule:

$latex \frac{{{x}^{-3}}{{y}^2}}{{{y}^2}{{x}^2}}=\frac{{{y}^2}}{{{y}^2}{{x}^2}{{x}^3}}$

Now, we apply the quotient rule to the variable y and the product rule to the variable x:

$latex \frac{{{y}^2}}{{{y}^2}{{x}^2}{{x}^3}}=\frac{{{y}^{2-2}}}{{{x}^{2+3}}}$

$latex =\frac{1}{{{x}^5}}$

Here, we can start with the negative exponents. Therefore, we take the reciprocal of the bases and change the exponents to positive:

$latex {{(7{{a}^3})}^{-2}}{{b}^{-1}}=\frac{1}{{{(7{{a}^3})}^2}{{b}^1}}$

Now, we apply the power of a power:

$latex \frac{1}{{{(7{{a}^3})}^2}{{b}^1}}=\frac{1}{{{7}^2}{{a}^6}{{b}^1}}$

$latex =\frac{1}{49{{a}^6}b}$

We start with the negative exponent on the right:

$latex {{({{b}^{-3}}c)}^2}\times {{({{b}^{2}}{{c}^3})}^{-3}}=\frac{{{({{b}^{-3}}c)}^2}}{{{({{b}^{2}}{{c}^3})}^3}}$

Now, we apply the power of a power rule to remove the parentheses:

$latex \frac{{{({{b}^{-3}}c)}^2}}{{{({{b}^{2}}{{c}^3})}^3}}=\frac{{{b}^{-6}}{{c}^2}}{{{b}^6}{{c}^9}}$

Again, we apply the negative exponents rule:

$latex \frac{{{b}^{-6}}{{c}^2}}{{{b}^6}{{c}^9}}=\frac{{{c}^2}}{{{b}^6}{{b}^6}{{c}^9}}$

Now, we apply the quotient rule to c and the product rule to b:

$latex \frac{{{c}^2}}{{{b}^6}{{b}^6}{{c}^9}}=\frac{1}{{{b}^{6+6}}{{c}^{9-2}}}$

$latex =\frac{1}{{{b}^{12}}{{c}^7}}$