Question

If $a^2=\log x$, $b^3=\log y$ and $3a^2-2b^3=6\log z$, then y in terms of x and z is $y=\frac{x^{\frac{3}{2}}}{z^m}$. The value of $m$ is equal to

• A. $2$
• B. $9$
• C. $0$
• D. $3$

Answer 1

The correct option is D $3$

$a^2=\log x{b}^3=\log y{3a}^2-2b^3=6\log z6\log z=3\log x-2\log y2\log y=3\log x-6\log z2\log y=\log \frac{x^3}{z^6}\log y^2=\log \frac{x^3}{z^6}\text{Removing log}{y}^2=\frac{x^3}{z^6}=\frac{x^{\frac{3}{2}}}{z^3}$