Alg9.4.40

If $\log_{10}(x^2-x-6)=x+\log_{10}(x+2)-4\,$ ,then find the value of x.

Taking the log values to one side,we get

$\log_{10}(x^2-x-6)-\log_{10}(x+2)=x-4\,$

$\log_{10}(\frac{(x^2-x-6)}{(x+2)})=x-4\,$

Factoring the numerator,we get

$\log_{10}(\frac{(x-3)(x+2)}{x+2})=x-4\,$

$\log_{10}(x-3)=x-4\,$

Taking out the logarithm,we get

$x-3=10^{x-4}\,$