for what value of a does $\\left(\\frac{1}{7}\\right)^{3a + 3}=343^{a - 1}$?\n-1\n0\n1\nno solution

Answer

Explanation:

Step1: Rewrite bases as powers of 7

We know that $\frac{1}{7}=7^{- 1}$ and $343 = 7^{3}$. So the equation $\left(\frac{1}{7}\right)^{3a + 3}=343^{a - 1}$ becomes $(7^{-1})^{3a + 3}=(7^{3})^{a - 1}$.

Step2: Apply power - of - a - power rule

According to the power - of - a - power rule $(x^{m})^{n}=x^{mn}$. So, $7^{(-1)\times(3a + 3)}=7^{3\times(a - 1)}$, which simplifies to $7^{-3a-3}=7^{3a - 3}$.

Step3: Set exponents equal

Since the bases are the same ($7$), we can set the exponents equal to each other: $-3a-3 = 3a-3$.

Step4: Solve for a

Add $3a$ to both sides: $-3=3a - 3+3a$, which simplifies to $-3 = 6a-3$. Then add 3 to both sides: $0 = 6a$. Divide both sides by 6, we get $a = 0$.