Question

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

(a) Rewrite the equation using the same base.
(b) Solve for x. Write your answer as a fraction in simplest form.

Answer 1

`\sf 3^(-3x)=3^((-4x+6))`

`\sf x=6`

Explanation:

`\sf Given \ equation: \left((1)/(27)\right)^x=3^((-4x+6))`

`\sf As \ (1)/(27)=(1)/(3^3) \ and \ (1)/(a^b)=a^(-b) \ then \ (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`