Question

Find the value of the unknown constant . The sum is in the picture. pls solve it in a notebook . it's urgent​

Answer 1

$\bf\red{Answer :-}$

c = 5

$\bf\red{Given :-}$

$\bigg( \frac{7}{9} \bigg) {}^{ - 3} \div \bigg( \frac{9}{7} \bigg) {}^{ - 2} \times \bigg( \frac{7}{9} \bigg) {}^{3c - 2} = \bigg( \frac{9}{7} \bigg) {}^{ - 8}$

$\bf\red{To\: find :-}$

Value of C

$\bf\red{Formulae\: implemented:- }$

From exponent and laws,

$\bigg( \frac{a}{b} \bigg) {}^{ - n} = \bigg( \frac{b}{a} \bigg) {}^{n}$

$a {}^{m} \times a {}^{n} = a {}^{m + n}$

$\bigg( \frac{a {}^{m} }{a {}^{n} } \bigg) = a {}^{m - n}$

If bases are equal then powers also equal.

$\bf\red{Solution:-}$

》 We shall first convert into same bases by using formula (a/b)⁻ⁿ = (b/a)ⁿ

》 Converting as a same base of 7/9 In L.H.S and R.H.S

$= \bigg( \frac{7}{9} \bigg) {}^{ - 3} \div \bigg( \frac{7}{9} \bigg) {}^{2} \times \bigg( \frac{7}{9} \bigg) {}^{3c - 2} = \bigg( \frac{7}{9} \bigg) {}^{8}$

Now all bases are same

$\frac{ \bigg( \frac{7}{9} \bigg) {}^{ - 3} \times \bigg( \frac{7}{9} \bigg) {}^{3c - 2} }{ \bigg( \frac{7}{9} \bigg) {}^{2} } = \bigg( \frac{7}{9} \bigg) {}^{8}$

Applying the 3rd formulae then denominator gets removed

$\bigg( \dfrac{a {}^{m} }{a {}^{n} } \bigg) = a {}^{m - n}$

$\bigg( \frac{7}{9} \bigg) {}^{ - 3 - 2} \times \bigg( \frac{7}{9} \bigg) {}^{3c - 2} = \bigg( \frac{7}{9} \bigg) {}^{8}$

$\bigg( \frac{7}{9} \bigg) {}^{ - 5} \times \bigg( \frac{7}{9} \bigg) {}^{3c - 2} = \bigg( \frac{7}{9} \bigg) {}^{8}$

Applying the 2nd formula

$a {}^{m} \times a {}^{n} = a {}^{m + n}$

$\bigg( \frac{7}{9} \bigg) {}^{ - 5 + 3c - 2 = } \bigg( \frac{7}{9} \bigg) {}^{8}$

$\bigg( \frac{7}{9} \bigg) {}^{ 3c - 7= } \bigg( \frac{7}{9} \bigg) {}^{8}$

As bases are equal Powers also should be equal So,

$3c - 7 = 8$

$3c = 7 + 8$

$3c = 15$

$c = 15 \div 3$

$\red{c = 5}$

$\bf\red{So, \:the\: value\: of\: c\: is\: 5}$