Solving this logarithm equation?

How do I solve this equation using common logarithms?

$\log x = 1-\log(x-3)$

• Can you write $1$ as $\log a$ for some $a$? – egreg Oct 18 '13 at 23:25
• Many mathematicians (not me) use "$\log$" for the natural logarithm (if you haven't heard of it, that is the logarithm using the base that is best suited for calculus). I always write $\log_{10}$ if I mean base $10$. That way there can be no confusion. – Stefan Smith Oct 19 '13 at 1:41

Hint: $$1 = \log x + \log(x - 3) \implies 1 = \log(x(x - 3))$$

Now use an exponent to remove the logarithm.

• $\log a+\log b=\log(ab)$ ;-) – egreg Oct 18 '13 at 23:28
• @egreg Whoops. Thanks! – user61527 Oct 18 '13 at 23:28

log(x) + log(x-3) =1

use log property

log(x . (x-3) ) = 1 eliminate log by writing the RHS as 10^1

x . (x-3) = 10^1

From here its expansion and solve the quadratic equation

Let $1 = \log 10$ then, $$\log x = \log 10 - \log (x - 3).$$ For all $x$ and $y$ and a constant $a$, one always has $$\log_a xy = \log_a x + \log_a y.\tag1$$ This is one of the most important logarithm rules. From this rule, we can express $\log\dfrac xy$. $$\log \frac xy = \log x\bigg(\frac 1y\bigg),$$ and from $(1)$, we see that $$\log \frac xy = \log x + \log \frac 1y = \log x + \log y^{-1}.$$ Let $\log y^{-1} = d$ then $10^d = y^{-1}\Leftrightarrow \dfrac{1}{10^d} = \dfrac{1}{y^{-1}} \Leftrightarrow 10^{-d} = y$. \begin{align}\therefore \log y &= -d \\ \Leftrightarrow -\log y &= d.\end{align} Since we asserted that $d = \log y^{-1}$ then, $$\log\frac xy = \log x - \log y.$$ Do you notice how useful the rule $(1)$ is? This implies that \begin{align} \log x &= \log 10 - \log (x - 3) \\ &\ \vdots \\ \Leftrightarrow x^2 - 3x - 10 &= 0. \end{align} We can solve for $x$ in this trinomial using the quadratic formula, such that \begin{align} x &= \frac{3\pm 7}{2} \\ \Leftrightarrow x = -2 \ &\text{ or } \ x = 5.\end{align} If you are not familiar with the quadratic formula, comment below.