Question

if a=b2x, b=c2y, c= a2z p.t xyz= 1/8

Answer 1

Appropriate Question :-

$\rm \: If \: a = {b}^{2x}, \: b = = {c}^{2y}, \: c = {a}^{2z}, \:prove \: that \: xyz = \dfrac{1}{8}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: a = {b}^{2x} - - - - (1)$

$\rm \: b = {c}^{2y} - - - - (2)$

$\rm \: c = {a}^{2z} - - - - (3)$

On substituting the value of a from equation (1) in (3), we get

$\rm \: c = {( {b}^{2x})}^{2z}$

We know,

$\boxed{\sf{ \:\rm \: {( {a}^{x} )}^{y} = {a}^{xy} \: }}$

So, using this identity, we get

$\rm \: c = {b}^{4xz}$

On substituting the value of b from equation (2), we get

$\rm \: c = {( {c}^{2y})}^{4xz}$

$\rm \: c = {c}^{8xyz}$

can be rewritten as

$\rm \: {c}^{1} = {c}^{8xyz}$

$\rm\implies \:\rm \: 8xyz = 1$

$\rm\implies \:\rm \: xyz = \dfrac{1}{8}$

Hence, Proved

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Additional Information

$\boxed{\sf{ \: {a}^{x} \times {a}^{y} = {a}^{x + y} \: }}$

$\boxed{\sf{ \: {a}^{x} \div {a}^{y} = {a}^{x - y} \: }}$

$\boxed{\sf{ \: {a}^{0} = 1 \: }}$

$\boxed{\sf{ \: {a}^{ - x} = \frac{1}{ {a}^{x} } \: }}$

$\boxed{\sf{ \: {a}^{x} \times {b}^{x} = {(a \times b)}^{x} \: }}$

$\boxed{\sf{ \: {a}^{x} = {a}^{y} \: \: \rm\implies \:x = y \: }}$

Answer 2

Step-by-step explanation:

Solution.

Given $\tt a=b^{2 x} \ldots(i) \qquad b=c^{2 y} \qquad \ldots (ii) \qquad c=a^{2 z}\qquad \ldots (iii) \qquad$ Substituting the value of b from (ii) in (i), we get

$\tt \: \[ a=\left(c^{2 y}\right)^{2 x}=c^{4 x y} \]$

Substituting the value of c from (iii) in (iv), we get

$\[ \begin{aligned} \tt a & \tt=\left(a^{2 x}\right)^{4 x y}=a^{8 x y z} \\ \Rightarrow & \tt \: a^{1} =a^{8 x y z} \Rightarrow \tt1=8 x y z \\ \Rightarrow & \tt x y z =\frac{1}{8} \end{aligned} \]$

(Assume $\sf a>0, a \neq 1$ )