Maths Question:
If ${{p}^{2}}=qr$ show that ${{\log }_{q}}p+{{\log }_{r}}p=2{{\log }_{q}}p{{\log }_{r}}p$
Maths Solution:
$\begin{align} & {{p}^{2}}=qr \\ & \text{Converting from indices to logarithms} \\ & {{\log }_{p}}qr=2 \\ & {{\log }_{p}}q+{{\log }_{p}}r=2 \\ & \frac{1}{{{\log }_{q}}p}+\frac{1}{{{\log }_{r}}p}=2\text{ Changing bases to }q\text{ and }r\text{ respectively lo}{{\text{g}}_{a}}b=\frac{1}{{{\log }_{b}}a} \\ & \frac{{{\log }_{r}}p+{{\log }_{q}}p}{{{\log }_{q}}p{{\log }_{r}}p}=2 \\ & {{\log }_{r}}p+{{\log }_{q}}p=2{{\log }_{q}}p{{\log }_{r}}p \\ & {{\log }_{q}}p+{{\log }_{r}}p=2{{\log }_{q}}p{{\log }_{r}}p \\\end{align}$
University mathstopic: