Question

If we have an algebraic equation as ${{a}^{x}}={{b}^{y}}={{c}^{z}}$ and ${{b}^{2}}=ac$ , Then the value of y is equal to:
(a) $\dfrac{2xz}{x+z}$
(b) $\dfrac{xz}{2\left( x-z \right)}$
(c) $\dfrac{xz}{2\left( z-x \right)}$
(d) $\dfrac{2xz}{x-z}$

Answer 1

Hint: Use the equation ${{a}^{x}}={{b}^{y}}={{c}^{z}}$ to get the value of a and b in terms of c, for instance ${{a}^{x}}={{c}^{z}}\Rightarrow a={{c}^{\dfrac{z}{x}}}$ . Now substitute the values of a and b found in the equation ${{b}^{2}}=ac$ and solve the equation you get to reach the answer.

Complete step-by-step solution -
It is given that ${{a}^{x}}={{b}^{y}}={{c}^{z}}$ . Using this we can say that ${{a}^{x}}={{c}^{z}}\text{ and }{{\text{b}}^{y}}={{c}^{z}}$ . If we use this equations to get the values of a and b in terms of c, we get
${{a}^{x}}={{c}^{z}}$
Now if we use the identity that ${{k}^{m}}=n\Rightarrow k={{n}^{\dfrac{1}{m}}}$ , we get
$a={{c}^{\dfrac{z}{x}}}......(i)$
Similarly, we can also get:
${{\text{b}}^{y}}={{c}^{z}}$
$\Rightarrow \text{b}={{c}^{\dfrac{z}{y}}}......(ii)$
Now we will move to the other equation given in the question.
${{b}^{2}}=ac$
Now, if we substitute the values of a and b from equation (i) and equation (ii), we will get\
${{c}^{\dfrac{2z}{y}}}={{c}^{\dfrac{z}{x}}}c$
Now, we know that ${{k}^{l}}{{k}^{m}}={{k}^{l+m}}$ . So, if we use this in our equation, we get
${{c}^{\dfrac{2z}{y}}}={{c}^{\dfrac{z}{x}+1}}$
Now we will take log of both the sides of the equation and use the identity $\log {{k}^{m}}=m\log k$ , we get
$\log {{c}^{\dfrac{2z}{y}}}=\log {{c}^{\dfrac{z}{x}+1}}$
$\Rightarrow \dfrac{2z}{y}\log c=\left( \dfrac{z}{x}+1 \right)\log c.........(iii)$
If we cancel logc from both sides of the equation, we get
$\dfrac{2z}{y}=\dfrac{z}{x}+1$
$\Rightarrow \dfrac{2z}{y}=\dfrac{z+x}{x}$
Now we will take the reciprocal of both the sides of the equation. On doing so, we get
$\dfrac{y}{2z}=\dfrac{x}{x+z}$
$\Rightarrow y=\dfrac{2xz}{x+z}$
Therefore, the answer to the above question is option (a).

Note: If you have noticed equation (iii), you will find that y can have other values as well when c=1, as if c=1 there is no constraint on y, as both sides of equation (iii) is zero, but as we have to decide according to the options, we took c to be not equal to one and solved. Also, we have considered a, b and c to be positive, else we cannot take log on both sides for getting equation (iii), as for taking log both sides of the equation must be positive.