Question

please help!!!!!!!!!

Answer 1

Answer:

Given :

Equation

$( \frac{4}{9} ) {}^{4} \times ( \frac{4}{9} ) {}^{ - 7} = ( \frac{4}{9} ) {}^{2x - 1}$

What to find out = value of x ?

Formulae used =

$1) \: \: a {}^{m} \times a {}^{n} = a {}^{m + n} \\ 2) \: \: a {}^{m} = a {}^{n} \\ = m = n$

Solution :-

$( \frac{4}{9} ) {}^{4} \times ( \frac{4}{9} ) {}^{ - 7} = ( \frac{4}{9} ) {}^{2x - 1} \\ = ( \frac{4}{9} ) {}^{ - 3} = ( \frac{4}{9} ) {}^{2x - 1} \\ now \: compare \: the \: power \\ - 3 = 2x - 1 \\ = 2x - 1 + 3 = 0 \\ = 2x + 2 = 0 \\ =x= \frac{ - 2}{2} \\ = x = - 1 \: \: answer$

Hence the value of X = - 1

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

Find the value of x

$\sf(\dfrac{4}{9})^4 \times(\dfrac{4}{9})^{-7} = (\dfrac{4}{9})^{2x-1}$

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

$\sf\implies(\dfrac{4}{9})^4 \times(\dfrac{4}{9})^-7 = (\dfrac{4}{9})^{2x-1}$

• $x^a \times x^b = x^{a+b}$

$\sf\implies(\dfrac{4}{9})^{4+(-7)} = (\dfrac{4}{9})^{2x-1}$

$\sf\implies(\dfrac{4}{9})^{-3} = (\dfrac{4}{9})^{2x-1}$

• when bases are equal then power will also be equal

$\sf\implies - 3 = 2x - 1$

$\sf\implies - 3 + 1= 2x$

$\sf\implies - 2 = 2x$

$\sf\implies x = \dfrac{-2}{2}$

$\sf\implies x = -1$

${\bold{\blue{\boxed{\bf{x = -1 }}}}}$

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