Jambmaths question:
If ${{\log }_{10}}(x-3)+{{\log }_{10}}(x-2)=lo{{g}_{10}}(2{{x}^{2}}-8)$, find the value of x
Option A:
7
Option B:
4
Option C:
–4
Option D:
–7
Jamb Maths Solution:
$\begin{align} & {{\log }_{10}}(x-3)+{{\log }_{10}}(x-2)=lo{{g}_{10}}(2{{x}^{2}}-8) \\ & {{\log }_{10}}(x-3)(x-2)={{\log }_{10}}(2{{x}^{2}}-8) \\ & \text{Equating both sides of the log}\text{.} \\ & (x-3)(x-2)=(2{{x}^{2}}-8) \\ & {{x}^{2}}-5x+6=2{{x}^{2}}-8 \\ & {{x}^{2}}+5x-14=0 \\ & (x+7)(x-2)=0 \\ & x=-7\text{ }or\text{ }x=2 \\\end{align}$
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