Logarithms

Algebra Level 3

$\large \frac {2\log \left( \frac {x-3}{x-7}\right)}{\log 3}+1= \frac {\log \left(\frac {x-3}{x-1}\right)}{\log 3}$

How many solutions does the equation above have?

1 negative and 1 positive solutions 2 positive solutions 2 negative solutions Only 1 negative solution

1 solution

Chew-Seong Cheong
Aug 19, 2017

$\begin{aligned} \frac {2\log \left( \frac {x-3}{x-7}\right)}{\log 3}+1 & = \frac {\log \left(\frac {x-3}{x-1}\right)}{\log 3} & \small \color{#3D99F6} \text{Multiply both sides by }\log 3 \\ 2\log \left( \frac {x-3}{x-7}\right) + \log 3 & = \log \left(\frac {x-3}{x-1}\right) \\ 2\log (x-3) - 2\log (x-7) + \log 3 & = \log (x-3) - \log(x-1) \\ \log (x-3) + \log (x-1) + \log 3 & = 2\log(x-7) \\ 3(x-3)(x-1) & = (x-7)^2 \\ 3x^2 - 12x+9 & = x^2 -14x + 49 \\ 2x^2 +2x - 40 & = 0 \\ x^2 + x - 20 & = 0 \\ (x+5)(x-4) & = 0 \end{aligned}$

We note that when $x=4$ , $\log\left(\frac {x-3}{x-7}\right) < 0$ is undefined. Therefore, the only solution is $x=-5$ and the answer is "only 1 negative solution" .

log (x - 7) being undefined is not the problem with x=4. The real problem is that the fraction

$\frac {x-3}{x-7} \text { is negative when x=4, hence } \log ( \frac {x-3}{x-7} ) \text { is undefined.}$

If we had e.g. (x - 8) in the numerator instead of (x - 3), then the fraction (of two negative numbers) would be positive, so the log of the fraction would be defined at x=4.

Zee Ell - 3 years, 9 months ago

Thanks. I have amended my solution.

Chew-Seong Cheong - 3 years, 9 months ago

Logarithms

1 solution

0 pending reports