Question

Here are a few basic facts on the first six Mughal emperors:
Babur (AD 1526-1530)
Humayun (AD 1530-1556)
Akbar (AD 1556-1605)
Jahangir (AD 1605-1627)
Shah Jahan (AD 1628-1658)
Aurangzeb (Alamgir) (AD 1658-1707)

Answer 1

Thank u It'll help me a lot

Answer 2

Explanation:

since, base are common ; i.e, (3) is common and they are multiplying, so there exponents will add such that :

\bf = > \frac{\red{( {3})^{2n + 2 + n} - {(3)}^{3n}}}{\red{({3})^{m} \times ({2})^{3} }} = \frac{\red{1}}{\red{27}} \\ \\ \bf = > \frac{\red{ {(3)}^{3n + 2} - {(3)}^{3n} }}{ \red{{(3)}^{3m} \times {(2)}^{3} } } = \frac{\red{1}}{\red{27} }

now , take {(3)}^{3n} common from numerator!

\bf = > \frac{\red{ {(3)}^{3n}( {3}^{2} - 1) }}{ \red{{(3)}^{3m} \times {(2)}^{3} }} = \frac{\red{1}}{\red{27}} \\ \\ \bf = > \frac{ \red{{(3)}^{3n}(9 - 1) }}{\red{ {(3)}^{3m} \times 8} } = \frac{\red{1}}{\red{27}} \\ \\ \bf = > \red{\frac{ {(3)}^{3n} \times \cancel8 }{ {(3)}^{3m} \times \cancel8}} = \red{\frac{1}{27} }

then, again (3) is common, but it is dividing, so exponents will subtract :

\bf = > \red{{(\cancel3)}^{3n - 3m}} = \red{( {\cancel3)}^{ - 3}} \\ \\ \bf= >\red{ 3n - 3m = - 3}

take 3 as common from LHS ,

\bf = > \red{3(n - m) = - 3}

take 3 to RHS from LHS , as it's multiplying, so taking other side, it will divide ;

\bf = > \red{n - m = - \frac{3}{3}} \\ \\ \bf = >\red{ n - m = - 1} \\ \\ \bf = > \red{\cancel - (m - n) = \cancel- 1} \\ \\ \bf = >\blue{\boxed{\huge{m - n = 1}}}

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