How do you solve log_6 5-log_6(x-7)=1?

Sep 18, 2016

$x = \frac{47}{6}$

Explanation:

Two principles at work here...

If you are subtracting the logs, divide the numbers.

All numbers or all logs, not both! Change 1 to a log.

${\log}_{6} 5 - {\log}_{6} \left(x - 7\right) = 1$

${\log}_{6} \left(\frac{5}{x - 7}\right) = {\log}_{6} 6 \text{ } \leftarrow$ if log A = log B, then A=B

$\frac{5}{x - 7} = 6$

$5 = 6 \left(x - 7\right)$

$5 = 6 x - 42$

$47 = 6 x$

$\frac{47}{6} = x$

Sep 18, 2016

$x = \frac{47}{6}$

Explanation:

Using the $\textcolor{b l u e}{\text{laws of logarithms}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\log x - \log y = \log \left(\frac{x}{y}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow {\log}_{6} 5 - {\log}_{6} \left(x - 7\right) = {\log}_{6} \left(\frac{5}{x - 7}\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\log}_{b} a = n \Leftrightarrow a = {b}^{n}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow {\log}_{6} \left(\frac{5}{x - 7}\right) = 1 \Rightarrow \frac{5}{x - 7} = {6}^{1} = 6$

$\Rightarrow 6 \left(x - 7\right) = 5 \Rightarrow 6 x - 42 = 5$

$\Rightarrow 6 x = 47 \Rightarrow x = \frac{47}{6}$

Sep 18, 2016

$x = \frac{47}{6}$

Explanation:

We have: ${\log}_{6} \left(5\right) - {\log}_{6} \left(x - 7\right) = 1$

Using the laws of logarithms:

$\implies {\log}_{6} \left(\frac{5}{x - 7}\right) = 1$

$\implies \frac{5}{x - 7} = {6}^{1}$

$\implies \frac{5}{x - 7} = 6$

$\implies 5 = 6 \left(x - 7\right)$

$\implies x - 7 = \frac{5}{6}$

$\implies x = \frac{47}{6}$

Therefore, the solution to the equation is $x = \frac{47}{6}$.