Question

Help me solve the equation for precalcules please and thank you

Answer 1

343^(3-2p) = 7^2

343 = 7^3

Hence 343^(3-2p) = 7^2

7^3(3-2p) = 7^2

Equating the base

3(3-2p) = 2

expand

9 - 6p = 2

subtract 9 from both sides

-6p = 2 - 9

-6p = -7

Divide both sides by -6

p = -7/-6

p = 7/6

p = 1.167