Question

please please answer my question.
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Answer 1

Question -

$If \: \: \frac{ {3}^{2x - 8} }{225} = \frac{ {5}^{3} }{ {5}^{x} } \: \: then \: find \: the \: value \: of \: x$

Solution -

$\implies \frac{ {3}^{2x - 8} }{225} = \frac{ {5}^{3} }{ {5}^{x} }$

$\implies \frac{ {3}^{2x - 8} }{ {3}^{2} \times {5}^{2} } = \frac{ {5}^{3} }{ {5}^{x} }$

$\implies \frac{ {3}^{2x - 8} }{ {3}^{2} \times {5}^{2} } = {5}^{3 - x}$

If

$\implies {3}^{2x - 8} \times {5}^{ - 2} = {5}^{3 - x} \times {3}^{2}$

Then,

$\implies {3}^{2x - 8} = {3}^{2}$

$\implies2x - 8 = 2$

$\implies2x = 2 + 8$

$\implies2x = 10$

$\implies \: x = \frac{10}{2}$

$\implies \: x = 5$

And,

$\implies {5}^{ - 2} = {5}^{3 - x}$

$\implies - 2 = 3 - x$

$\implies - 2 - 3 = - x$

$\implies - 5 = - x$

$\implies \: x = 5$

Therefore the value of x is 5

Laws of exponents-

$1) {a}^{m} \times {a}^{n} = {a}^{m + n}$

$2) ({a}^{m} ) {}^{n} = {a}^{mn}$

$3) \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$4) \frac{1}{ {a}^{m} } = {a}^{ - m}$

$5) {a}^{m} \times {b}^{m} = ( {ab})^{m}$

$6) {a}^{m} = {a}^{n} \implies \: m = n$