Question

Tìm x

\(\dfrac{\left(-2\right)^{x-1}}{18}=\dfrac{8}{9}\)

Answer 1

\(\dfrac{\left(-2\right)^{x-1}}{18}=\dfrac{8}{9}\)

\(\Leftrightarrow x-1=4\)

hay x=5

Answer 2

\(\dfrac{\left(-2\right)^{x-1}}{18}=\dfrac{8}{9}=\dfrac{16}{18}\)

⇒\(\left(-2\right)^{x-1}=8\)

⇒\(\left(-2\right)^{x-1}=2^3\)

vì 2^3 >0 mà (-2)^3<0

⇒x không tồn tại

Answer 3

\(\dfrac{\left(-2\right)^{x-1}}{18}=\dfrac{8}{9}=\dfrac{16}{18}\Leftrightarrow\left(-2\right)^{x-1}=16=\left(-2\right)^4\Leftrightarrow x-1=4\Leftrightarrow x=5\)

Answer 4

\(\dfrac{\left(-2\right)^{x-1}}{18}=\dfrac{8}{9}\)

<=> \(\dfrac{\left(-2\right)^{x-1}}{18}:\dfrac{8}{9}=1\)

<=> \(\dfrac{\left(-2\right)^{x-1}}{18}.\dfrac{9}{8}=1\)

<=> \(\dfrac{2^{x-1}-1}{2.2^3}=1\)

<=> \(\dfrac{2^{x-1}}{2^4}=1\)

<=> 2^{x - 1 - 4} = 1

<=> 2^{x - 5} = 1

<=> x = 5