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How do you solve $log (x - 3) + log x = 1$ $log(x-3)+log x=1$ ?

1 Answer

Bdub · EZ as pi

May 4, 2016

$x = 5$ $x=5$

Explanation:

Use Properties: ${log}_{b} (x y) = {log}_{b} x + {log}_{b} y$ $log_b(xy)=log_b x+log_by$

${log}_{b} x = y \Leftrightarrow b^{y} = x$ $log_bx=y iff b^y=x$

$log (x (x - 3)) = 1$ $log (x(x-3))=1$ $\times \times \times$ $color(white)(xxxxxx)$ [ 1 = log10]

$log (x^{2} - 3 x) = log 10$ $log(x^2-3x)=log10$

$x^{2} - 3 x^{1} = 10^{1}$ $x^2-3x^1 = 10^1$

$x^{2} - 3 x - 10 = 0$ $x^2-3x-10=0$

$(x - 5) (x + 2) = 0$ $(x-5)(x+2)=0$

$x = 5 or x = - 2$ $x=5 or x=-2$

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