# Solving this logarithm equation? How do I solve this equation using

Solving this logarithm equation?
How do I solve this equation using common logarithms?
$\mathrm{log}x=1-\mathrm{log}\left(x-3\right)$
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Mekjulleymg
Hint:
$1=\mathrm{log}x+\mathrm{log}\left(x-3\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}1=\mathrm{log}\left(x\left(x-3\right)\right)$
Now use an exponent to remove the logarithm.

Janet Forbes
Let $1=\mathrm{log}10$ then,
$\mathrm{log}x=\mathrm{log}10-\mathrm{log}\left(x-3\right).$
For all x and y and a constant a, one always has
$\begin{array}{}\text{(1)}& {\mathrm{log}}_{a}xy={\mathrm{log}}_{a}x+{\mathrm{log}}_{a}y.\end{array}$
This is one of the most important logarithm rules. From this rule, we can express $\mathrm{log}\frac{x}{y}$
$\mathrm{log}\frac{x}{y}=\mathrm{log}x\left(\frac{1}{y}\right),$
and from (1), we see that
$\mathrm{log}\frac{x}{y}=\mathrm{log}x+\mathrm{log}\frac{1}{y}=\mathrm{log}x+\mathrm{log}{y}^{-1}.$
${10}^{d}={y}^{-1}⇔\frac{1}{{10}^{d}}=\frac{1}{{y}^{-1}}⇔{10}^{-d}=y$
$\begin{array}{rl}\therefore \mathrm{log}y& =-d\\ ⇔-\mathrm{log}y& =d.\end{array}$
Since we asserted that $d=\mathrm{log}{y}^{-1}$ then
$\mathrm{log}\frac{x}{y}=\mathrm{log}x-\mathrm{log}y.$
Do you notice how useful the rule (1) is? This implies that

We can solve for x in this trinomial using the quadratic formula, such that

If you are not familiar with the quadratic formula, comment below.