Question: Solve: $10^{3y-3} = \dfrac{1}{0.001}$


Solution:
Given,

$10^{3y-3} = \dfrac{1}{0.001}$

$or, 10^{3y -3} = \dfrac{1}{\dfrac{1}{1000}}$

$or, 10^{3y -3} = \dfrac{1}{\dfrac{1}{10^3}}$

$or, 10^{3y -3} = \dfrac{1 * 10^3}{1}$

$or, 10^{3y -3} = 10^3$

$or, 3y -3 = 3$

$or, 3y = 3+3$

$or, 3y = 6$

$or, \dfrac{3y}{3} = \dfrac{6}{3}$

$\therefore, y = 2$
= Answer

Related Notes and Solutions:

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