Question

If a^x=b^y-c^z and b^2=ac . Prove that 1/x+1/z=2/y

Answer 1

Answer:

I hope it would help you

Answer 2

Answer:

$\large \text{ \dfrac{2}{y}=\dfrac{1}{x} +\dfrac{1}{z} Proved.}$

Step-by-step explanation:

Given :

$\large \text{ a^x=b^y-c^z and b^2=ac }\\\\\\\large \text{We have to prove \dfrac{1}{x} +\dfrac{1}{z}=\dfrac{2}{y} }\\\\\\\large \text{Taking log both side}\\\\\\\large \text{ log \ a^x=log \ b^y and log \ b^2=log \ ac }\\\\\\\large \text{Here we will use three formula of log}\\\\\\\large log \ mn=log \ m+log \ n\\\\\\\large log \ \dfrac{m}{n}=log \ m - log \ n\\\\\\\large log \ a^x=x \ log \ a$

$\large \text{ log \ a^x=log \ b^y-log \ c^z and log \ b^2=log \ ac }\\\\\\\large \text{ xlog \ a=ylog \ b-zlog \ c and 2log \ b=log \ ac }\\\\\\\large \dfrac{xlog \ a+zlog \ c}{y} =log \ b \ ....(i)\\\\\\\\\large log \ b=\dfrac{log \ ac}{2} \ .....(ii) \\\\\\\large \text{From (i) and (ii) we get }$

$\large \dfrac{xlog \ a+zlog \ c}{y}=\dfrac{log \ ac}{2} \\\\\\\large \dfrac{2}{y}=\dfrac{log \ ac}{xlog \ a+zlog \ c} \\\\\\\large \dfrac{2}{y}=\dfrac{log \ a+log \ c}{xlog \ a+zlog \ c} \\\\\\\large \text{From above we get }\\\\\\\large \dfrac{2}{y}=\dfrac{1}{x} +\dfrac{1}{z}$

Hence proved.