Question

Plz solve this question.......

6.Given log10 x = a, log10 y = b and log10 z = C,
(1) write down 1020-3 in terms of x.
(ii) write down 103b-1 in terms of y.
(iii) if log10 P = 2a +
3c, express P in terms of x, y and z.
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Answer 1

Answer:

$we \: know \: {10}^{a} = x \frac{1}{2}$

$10 \frac{b}{2} = y$

${10}^{b} = {y}^{2}$

$= > log^{p} 10 = 3a - 2b$

$= > p = {10}^{3} a - 2b$

$= > p = ( {10}^{3} {)}^{a} \div ( {10}^{2} )b$

$= > p = (10a {)}^{3} \div ( {10}^{b} {)}^{2}$

$substituting \: {10}^{a} \: and \: {10}^{b} we \: get$

$= > p = (x \frac{1}{2} {)}^{3} \div ( {y}^{2} )$

$= > p = x \frac{3}{2} \div {y}^{4}$

$= > p = {x}^{ \frac{3}{2} } \div {y}^{4}$

$= > p \frac{ {x}^{ \frac{3}{2} } }{ {y}^{4} }$

Hope it's help you......

Answer 2

Answer:

please see above attachment

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