Maths Question:
If ${{\log }_{3}}(x-6)=2y$ and ${{\log }_{2}}(x-7)=3y$ show that ${{x}^{2}}-13x+42={{72}^{y}}$ Given that y = 1. Find the possible value(s) of x
Maths Solution:
$\begin{align} & {{\log }_{3}}(x-6)=2y \\ & x-6={{3}^{2y}}----(i) \\ & x={{9}^{y}}+6---(i) \\ & {{\log }_{2}}(x-7)=3y \\ & x-7={{2}^{3y}}---(ii) \\ & Multiply(i)and(ii)\text{ together} \\ & (x-6)(x-7)={{3}^{2y}}\times {{2}^{3y}} \\ & {{x}^{2}}-13x+42={{9}^{y}}\times {{8}^{y}} \\ & {{x}^{2}}-13x+42={{72}^{y}} \\ & \text{Given that }y=1 \\ & {{x}^{2}}-13x+42=72 \\ & {{x}^{2}}-13x-30=0 \\ & (x-15)(x+2)=0 \\ & x=15\text{ or }x=-2 \\\end{align}$
University mathstopic: