Question

-1
0
1
no solution
Help asappp

Answer 1

Step-by-step answer:

Given:

(1/7)^(3a+3) = 343^(a-1) ................................................(1)

Find the value of a that satisfies equation (1)

Solution:

Using the law of exponents, (1/x)^p = (x)^(-p), and apply to LHS (left-hand side) of (1)

(7)^(-3a-3) = 343^(a-1) ...................................................(2)

We know that 7^3 = 343, substitute on the RHS (right-hand side) of (2)

(7)^(-3a-3) = (7^3)^(a-1) ...................................................(3)

Using the law of exponents, (x^p)^q = x^(pq) and apply to the RHS of (3)

(7)^(-3a-3) = (7)^(3(a-1)) = 7^(3a-3) ...................................................(3)

now that the bases on each side are equal, we apply the rule

x^(p) = x^(q) => p=q for all x>0, and apply to (3)

7^(-3a-3) = 7(3a-3) =>

-3a-3 = 3a-3............................................................................................(4)

Solve (4) for a

-3 + 3 = 3a+3a

0 = 6a

a=0..........................................................................................................(5)

Check:

Substitute a=0 in (1)

LHS : (1/7)^(0+3) = (1/7)^(3) = (7)^(-3)

RHS : 343^(0-1) = (7^3)^(-1) = (7)^(-3)

Since LHS equals RHS, the solution is validated.

Answer 2

0

Explanation:

As you know 7^3 = 343

So

(1/7)^(3a + 3) = 343^(a-1)

(7^-1)^(3a + 3) = (7^3)^(a-1)

7^(-3a - 3) = 7^(3a - 3)

Both sides have same base of 7

So

-3a - 3 = 3a - 3

-6a = 0

a = 0