Alg9.4.23

solution If $\frac{\log_2(x)}{4}=\frac{\log_2(y)}{6}=\frac{\log_2(z)}{3P},x^3y^2z=1\,$ then find the value of P.

Let

$\frac{\log_2(x)}{4}=\frac{\log_2(y)}{6}=\frac{\log_2(z)}{3P}=n\,$

Then

$\log_2(x)=4n,\log_2y=6n,\log_2(z)=3Pn\,$

Hence

$x=2^{4n},y=2^{6n},z=2^{3Pn}\,$

Substituting these values in $x^3y^2z=1\,$

$2^{12n}\cdot2^{12n}\cdot2^{3Pn}=2^{0}\,$

$2^{24n+3Pn}=0\,$

Hence P=-8.