Question

solve for x: (-3/19) to the power8 ×(-3/19) to the power - 18 =( - 3/19) to the power 3x-9 ​

Answer 1

Given:

The equation:-

$\rm {\bigg( \dfrac{-3}{19} \bigg)^{8} \times \bigg( \dfrac{-3}{19} \bigg)^{-18} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

What To Find:

We have to find the value of x.

Solution:

$\rm {\bigg( \dfrac{-3}{19} \bigg)^{8} \times \bigg( \dfrac{-3}{19} \bigg)^{-18} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

Solve the exponents in LHS,

⇒ $\rm {\bigg( \dfrac{-3}{19} \bigg)^{8 + (-18)} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

Add 8 and -18,

⇒ $\rm {\bigg( \dfrac{-3}{19} \bigg)^{-10} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

Since the bases are the same, we can omit them,

⇒ -10 = 3x - 9

Take 9 to LHS,

⇒ -10 - 9 = 3x

Subtract 9 from -10,

⇒ -19 = 3x

Take 3 to LHS,

⇒ $\rm \dfrac{-19}{3} = x$

∴ Hence, x = $\dfrac{-19}{3}$.

Answer 2

Given:

The equation:-

$\rm {\bigg( \dfrac{-3}{19} \bigg)^{8} \times \bigg( \dfrac{-3}{19} \bigg)^{-18} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

What To Find:

We have to find the value of x.

Solution:

$\rm {\bigg( \dfrac{-3}{19} \bigg)^{8} \times \bigg( \dfrac{-3}{19} \bigg)^{-18} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

Solve the exponents in LHS,

⇒ $\rm {\bigg( \dfrac{-3}{19} \bigg)^{8 + (-18)} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

Add 8 and -18,

⇒ $\rm {\bigg( \dfrac{-3}{19} \bigg)^{-10} = \bigg( \dfrac{-3}{19} \bigg)^{3x - 9}}$

Since the bases are the same, we can omit them,

⇒ -10 = 3x - 9

Take 9 to LHS,

⇒ -10 - 9 = 3x

Subtract 9 from -10,

⇒ -19 = 3x

Take 3 to LHS,

⇒ $\rm \dfrac{-19}{3} = x$

∴ Hence, x = $\dfrac{-19}{3}$.