If-16-a-b-2-2-16-b-c-2-16-a-c-2-2-then-c-

Question Number 132198 by benjo_mathlover last updated on 12/Feb/21

If {\begin{cases} 16^{a + b} = \frac{\sqrt{2}}{2} \\ 16^{b + c} = 2 \\ 16^{a + c} = 2 \sqrt{2} \end{cases}

then c =__

Answered by EDWIN88 last updated on 12/Feb/21

eq (1) 16^{2 (a + b)} = 2^{- 1} \Rightarrow 8 a + 8 b = - 1

eq (2) 16^{b + c} = 2 \Rightarrow 4 b + 4 c = 1

eq (3) 16^{a + c} = 2 \sqrt{2} = 2^{3 / 2} \Rightarrow 4 a + 4 c = \frac{3}{2}

eliminate eq (1) \land eq (2) \Rightarrow 8 a + 8 b = - 1

8 c + 8 b = 2

_________-

eq (4) 8 a - 8 c = - 3

8 a + 8 c = 3

___________-

- 16 c = - 6 \Rightarrow c = \frac{3}{8}

Answered by Rasheed.Sindhi last updated on 14/Feb/21

A n O t h e r w a y

{\begin{cases} 16^{a + b} = \frac{\sqrt{2}}{2} \dots \dots . . (i) \\ 16^{b + c} = 2 \dots \dots \dots . (i i) \\ 16^{a + c} = 2 \sqrt{2} \dots \dots (i i i) \end{cases}

(i) / (i i) :

16^{a - c} = \frac{\sqrt{2}}{4} \dots \dots \dots \dots \dots . (i v)

(i i i) / (i v) :

16^{2 c} = 2 \sqrt{2} . \frac{4}{\sqrt{2}} = 8

2^{8 c} = 2^{3}

c = 3 / 8