Question

Use the equation, (1/27)^x = 3^-4x+6, to complete the following problems

Answer 1

Answer:

$\sf 3^{-3x}=3^{(-4x+6)}$

$\sf x=6$

Step-by-step explanation:

$\sf Given \ equation: \left(\dfrac{1}{27}\right)^x=3^{(-4x+6)}$

$\sf As \ \dfrac{1}{27}=\dfrac{1}{3^3} \ and \ \dfrac{1}{a^b}=a^{-b} \ then \ \dfrac{1}{27}=3^{-3}$

Therefore, we can rewrite the given equation with base 3:

$\implies \sf (3^{-3})^x=3^{(-4x+6)}$

Apply the exponent rule $\sf (a^b)^c=a^{bc}$ :

$\implies \sf 3^{-3x}=3^{(-4x+6)}$

$\sf If \ a^{f(x)}=a^{g(x)} \ then \ f(x)=g(x)$

$\implies -3x=-4x+6$

Add 4x to both sides to solve for x:

$\implies \sf x=6$