Question

if p and q are positive integers such that p^q=q^p and q=27p, then find the value of p^13 q^1/9​

Answer 1

$p^{13}q^{\frac{1}{9}}=\sqrt[9]{3^{118}\cdot 2^2}$

Step-by-step explanation:

$q^p=p^q$

$q=27p$

Substitute the value of q

$(27p)^p=p^q$

$(3^3p)^p=p^q$

$(3)^{3p}p^p=p^q$

Using rule : $(ab)^x=a^xb^x$

$(3)^{3p}=\frac{p^q}{p^p}=p^{q-p}$

Using rule : $\frac{a^x}{a^y}=a^{x-y}$

On comparing both sides

Then, we get

$p=3, q-p=3p$

$q-p=3p$

$q=3p+p=4p$

Substitute the value of p

$q=4(3)=12$

Substitute the values of p and q

$p^{13}q^{\frac{1}{9}}=(3)^{13}(12)^{\frac{1}{9}}$

$p^{13}q^{\frac{1}{9}}=3^{13}(3\times 4)^{\frac{1}{9}}$

$p^{13}q^{\frac{1}{9}}=3^{13}\times 3^{\frac{1}{9}\times 4^{\frac{1}{9}}$

$p^{13}q^{\frac{1}{9}}=3^{13+\frac{1}{9}}\times (4)^{\frac{1}{9}}$

Using rule :$a^x\times a^y=a^{x+y}$

$p^{13}q^{\frac{1}{9}}=3^{\frac{118}{9}}\cdot (2^2)^{\frac{1}{9}}$

$p^{13}q^{\frac{1}{9}}=\sqrt[9]{3^{118}\cdot 2^2}$

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