Question

hw04b fall25: problem 5
(1 point)
$(x^{61}cdot y^{126})cdot(x^{24}) = x^{p}cdot y^{q}$ when $p = \square$ and $q = \square$
$(e^{61x}cdot(\sin(11x))^{126})cdot(e^{24x}) = e^{px}cdot(\sin(11x))^{q}$ when $p = \square$ and $q = \square$
$\frac{x^{99}cdot y^{121}}{x^{24}} = x^{p}cdot y^{q}$ when $p = \square$ and $q = \square$
$\frac{x^{24}cdot y^{121}}{x^{99}} = x^{p}cdot y^{q}$ when $p = \square$ and $q = \square$
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Explanation:

Step1: Use exponent - product rule

For the first equation \((x^{61}\cdot y^{126})\cdot(x^{24})=x^{P}\cdot y^{Q}\), when multiplying terms with the same base \(x\), we add the exponents. So \(P = 61 + 24=85\) and \(Q = 126\).

Step2: Apply exponent - product rule for exponential function

For the second equation \((e^{61x}\cdot(\sin(11x))^{126})\cdot(e^{24x})=e^{Px}\cdot(\sin(11x))^{Q}\), for the base \(e\), we add the exponents of \(e\). So \(P=61 + 24 = 85\) and \(Q = 126\).

Step3: Use exponent - quotient rule

For the third equation \(\frac{x^{99}\cdot y^{121}}{x^{24}}=x^{P}\cdot y^{Q}\), when dividing terms with the same base \(x\), we subtract the exponents. So \(P=99 - 24=75\) and \(Q = 121\).

Step4: Apply exponent - quotient rule again

For the fourth equation \(\frac{x^{24}\cdot y^{121}}{x^{99}}=x^{P}\cdot y^{Q}\), for the base \(x\), we subtract the exponents. So \(P=24-99=- 75\) and \(Q = 121\).

Answer:

First equation: \(P = 85\), \(Q = 126\)
Second equation: \(P = 85\), \(Q = 126\)
Third equation: \(P = 75\), \(Q = 121\)
Fourth equation: \(P=-75\), \(Q = 121\)