Question

If x, y, z are in G.P., $a^x=b^y=c^z$, then-

• A. $lo{g}_{c}b=lo{g}_{a}c$
• B. $lo{g}_{1}c=lo{g}_{b}c$
• C. $lo{g}_{a}b=lo{g}_{c}b$
• D. $lo{g}_{b}a=lo{g}_{c}b$

Answer 1

The correct option is A $lo{g}_{c}b=lo{g}_{a}c$

$\begin{array}{c}\frac{x(y+z-x)}{logx}=\frac{y(z+x-y)}{logy}=\frac{z(x+y-z)}{logz}=K\left(Let\right),\\ Now,weget----|\begin{array}{c}ifbaseisnotgiventhen,\\ applybaseas=10\end{array}\\ \\ lo{g}_{10}x=\frac{x(y+z-x)}{k}\Rightarrow x={10}^{\frac{x(y+z-x)}{k}}\\ {\log }_{10}y=\frac{y(z+x-y)}{k}\Rightarrow y={10}^{\frac{y(z+x-y)}{k}}\\ {\log }_{10}z=\frac{z(x+y-z)}{k}\Rightarrow z={10}^{\frac{z(x+y-z)}{k}}\\ Now,wefind{x}^y{y}^{x}isequalto......\\ so,\\ x^y={\left[{10}^{\frac{x(y+z-x)}{k}}\right]}^y\Rightarrow {10}^{\frac{xy(y+z-x)}{k}}\\ y{}^x={\left[{10}^{\frac{y(z+x-y)}{k}}\right]}^x\Rightarrow {10}^{\frac{xy(z+x-y)}{k}}\\ \\ \therefore x^y.x^y={10}^{\frac{xy}{k}(y+z-x+z+x-y)}={10}^{\frac{2xyz}{k}}-------\left(i\right)\\ \\ ifwetake,x^z{z}^x...\\ \\ x^z={\left[{10}^{\frac{x(y+z-x)}{k}}\right]}^z\Rightarrow {10}^{\frac{zx(y+z-x)}{k}}\\ z^x={\left[{10}^{\frac{z(x+y-z)}{k}}\right]}^x\Rightarrow {10}^{\frac{zx(x+y-z)}{k}}\\ \\ \therefore x^z.z^x={10}^{\frac{zx}{k}(y+z-x+x+y-z)}={10}^{\frac{2xyz}{k}}-------\left(ii\right)\\ So,(x^y.x^y)isequalto(x^z.z^x)andthecorrectoptionisA.\end{array}$