# extraneous roots with logs?

• Dec 14th 2009, 04:23 PM
Barthayn
extraneous roots with logs?
Hi, I have these four questions that I cannot seem to understand what they are asking me to solve. I do not know how to begin the question.

The questions are as followed:

3. Solve. Identify and reject any extraneous roots. Check your solutions using graphing technology.

a) $log x + log (x-4) = 1$

b) $log x^3 - log 2 = log (2x^2)$

c) $log (v-1) = 2 + log (v-16)$

d) $1 + log y = log (y+9)$

How would I do these? My teacher only showed us the way to solve these when they are adding? (Worried)
• Dec 14th 2009, 05:18 PM
Prove It
Quote:

Originally Posted by Barthayn
Hi, I have these four questions that I cannot seem to understand what they are asking me to solve. I do not know how to begin the question.

The questions are as followed:

3. Solve. Identify and reject any extraneous roots. Check your solutions using graphing technology.

a) $log x + log (x-4) = 1$

b) $log x^3 - log 2 = log (2x^2)$

c) $log (v-1) = 2 + log (v-16)$

d) $1 + log y = log (y+9)$

How would I do these? My teacher only showed us the way to solve these when they are adding? (Worried)

I assume these are natural logarithms? As in, base $e$?

1. $\log{x} + \log{(x + 4)} = 1$

$\log{[x(x + 4)]} = 1$

$x(x + 4) = e^1$

$x(x + 4) = e$

$x^2 + 4x = e$

$x^2 + 4x - e = 0$

2. $\log (x^3) - \log 2 = \log (2x^2)$

$\log{\frac{x^3}{2}} = \log{(2x^2)}$

$\frac{x^3}{2} = 2x^2$

$\frac{1}{3}x^3 - 2x^2 = 0$

$x^2\left(\frac{1}{3}x - 2\right) = 0$

Solve for $x$.

3. $\log (v-1) = 2 + \log (v-16)$

$\log{(v - 1)} - \log{(v - 16)} = 2$

$\log{\left(\frac{v - 1}{v - 16}\right)} = 2$

$\frac{v - 1}{v - 16} = e^2$

$\frac{v - 16 + 15}{v - 16} = e^2$

$1 + \frac{15}{v - 16} = e^2$

$\frac{15}{v - 16} = e^2 - 1$

$\frac{v - 16}{15} = \frac{1}{e^2 - 1}$

$v - 16 = \frac{15}{e^2 - 1}$

$v = \frac{15}{e^2 - 1} + 16$.

4. $1 + \log y = \log (y+9)$

$\log{(y + 9)} - \log{y} = 1$

$\log{\frac{y + 9}{y}} = 1$

$\frac{y + 9}{y} = e^1$

$\frac{y + 9}{y} = e$

$1 + \frac{9}{y} = e$

$\frac{9}{y} = e - 1$

$\frac{y}{9} = \frac{1}{e - 1}$

$y = \frac{9}{e - 1}$.