O lvl A Maths: Indices and Logarithms

(a) Given that $\left ( 144p^{4} \right )^{\frac{3}{2}}\div \left ( 216p^{-3} \right )^{-\frac{3}{2}} = 2^{x} 3^{y}p^{z}$ , evaluate x, y and z.

(b) Given that 2^2x+2 * 5^x-1 = 8^x * 5^2x, evaluate 10^x.

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Answer:
(a)
$\left ( 144p^{4} \right )^{\frac{3}{2}}\div \left ( 216p^{-3} \right )^{-\frac{3}{2}} = 2^{x} 3^{y}p^{z}$
$\left (16\times 9\times p^{4} \right )^{\frac{3}{2}}\div \left ( 27\times 8\times p^{-3} \right )^{-\frac{3}{2}} = 2^{x} 3^{y}p^{z}$
$\left (2^{4}\times 3^{2}\times p^{4} \right )^{\frac{3}{2}}\div \left ( 3^{3}\times 2^{3}\times p^{-3} \right )^{-\frac{3}{2}} = 2^{x} 3^{y}p^{z}$
$\left (2^{4\times \frac{3}{2}}\times 3^{2\times \frac{3}{2}}\times p^{4\times \frac{3}{2}} \right )\div \left ( 3^{3\times \left (-\frac{3}{2} \right )}\times 2^{3\times \left (-\frac{3}{2} \right )}\times p^{-3\times \left (-\frac{3}{2} \right )} \right ) = 2^{x} 3^{y}p^{z}$
$\left (2^{6}\times 3^{3}\times p^{6} \right )\div \left ( 3^{-\frac{9}{2}}\times 2^{-\frac{9}{2}}\times p^{\frac{9}{2}} \right ) = 2^{x} 3^{y}p^{z}$
$\left (2^{6-\left (-\frac{9}{2} \right )}\times 3^{3-\left (-\frac{9}{2} \right )}\times p^{6-\left (\frac{9}{2} \right )} \right ) = 2^{x} 3^{y}p^{z}$
$2^{10\frac{1}{2}}\times 3^{7\frac{1}{2}}\times p^{1\frac{1}{2}} = 2^{x} 3^{y}p^{z}$

By comparison,
x = 10 1/2
y = 7 1/2
z = 1 1/2 (ans)

(b)
2^2x+2 * 5^x-1 = 8^x * 5^2x

2^2x2² * 5^x5^-1 = (2³)^x * 5^2x
4/5 = 2^3x-2x * 5^2x-x
4/5 = 2^x * 5^x
10^x = 4/5 (ans)

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O lvl A Maths: Indices and Logarithms