Question

prove that x^(lny-lnz)×y^(lnz-lnx)×z^(lnx-lny)=1
​

Answer 1

$\textbf{To prove:}$

$x^{logy-logz}{\times}y^{logz-logx}{\times}z^{logx-logy}=1$

$\textbf{Solution:}$

$\textbf{Logarithmic notation:}$

$a^x=N\;\implies\;log_aN=x$

$\text{N can be written as,}\,N=a^{log_aN}$

$\text{Consider,}$

$x^{logy-logz}$

$=(e^{logx})^{logy-logz}$

$=e^{logx(logy-logz)}$

$=e^{logx\;logy-logx\;logz}$

$\implies\bf\,x^{logy-logz}=e^{logx\;logy-\logx\;logz}$

$\text{Similarly,}$

$y^{logz-logx}=e^{logy\;logz-logy\;logx}$

$z^{logx-logy}=e^{logz\;logx-logz\;logy}$

$\text{Now,}$

$x^{logy-logz}{\times}y^{logz-logx}{\times}z^{logx-logy}$

$=e^{logx\;logy-logx\;logz}{\times}e^{logy\;logz-logy\;logx}{\times}e^{logz\;logx-logz\;logy}$

$=e^{logx\;logy-logx\;logz+logy\;logz-logy\;logx+logz\;logx-logz\;logy}$

$=e^0$

$=1$

$\textbf{Answer:}$

$\boxed{\bf\,x^{logy-logz}{\times}y^{logz-logx}{\times}z^{logx-logy}=1}$