Question

For what value of a does (one-seventh) Superscript 3 a + 3 Baseline = 343 Superscript a minus 1?

Answer 1

Answer:

$a = 0$

Step-by-step explanation:

Given

$(\frac{1}{7})^{3a+3} = (343)^{a-1}$

Required

Find a

We start by writing 343 as a factor of 7

$(\frac{1}{7})^{3a+3} = (343)^{a-1}$

$(\frac{1}{7})^{3a+3} = (7^3)^{a-1}$

From laws of indices;

$\frac{1}{a} = a^{-1}$

So;

$(\frac{1}{7})^{3a+3} = (7^3)^{a-1}$ becomes

$(7^{-1})^{3a+3} = (7^3)^{a-1}$

$(7)^{(-1){(3a+3)}} = (7)^{3(a-1)}$

Cancel 7 on both sides

$(-1){(3a+3) = {3(a-1)}$

Open brackets

$-3a - 3 = 3a - 3$

Collect like terms

$3 - 3 = 3a + 3a$

$0 = 6a$

Divide both sides by 6

$0 = a$

$a = 0$