# Logarithm questions

• Jun 22nd 2007, 07:47 PM
kevinz
Logarithm questions
Solve the following logarithmic equations:
a) log2 (6x-8) - log2 (2x+4) = log2 (x-6)
b) log3 (x+3) - log3 (x-3) = 1
c) logn 1 = x

kinda stuck on those ones
• Jun 22nd 2007, 08:03 PM
Jhevon
Quote:

Originally Posted by kevinz
Solve the following logarithmic equations:
a) log2 (6x-8) - log2 (2x+4) = log2 (x-6)

$\displaystyle \log_2 (6x - 8) - \log_2 (2x + 4) = \log_2 (x - 6)$

$\displaystyle \Rightarrow \log_2 \left( \frac {6x - 8}{2x + 4} \right) = \log_2 (x - 6)$ ............since $\displaystyle \log_a x - \log_a y = \log_a \left( \frac {x}{y}\right)$

Now we can just drop the logs:

$\displaystyle \Rightarrow \frac {6x - 8}{2x + 4} = x - 6$

I think you can take it from here

Quote:

b) log3 (x+3) - log3 (x-3) = 1
$\displaystyle \log_3 (x + 3) - \log_3 (x - 3) = 1$

$\displaystyle \Rightarrow \log_3 \left( \frac {x + 3}{x - 3} \right) = 1$

$\displaystyle \Rightarrow 3^1 = \frac {x + 3}{x - 3}$ ...........since if $\displaystyle \log_a b = c$, then $\displaystyle a^c = b$

...and i think you can solve that equation

Quote:

c) logn 1 = x
$\displaystyle x = 0$ right off the bat!

The $\displaystyle \log_a 1 = 0$ always, i don't care what the base is, the log of 1 is 0, why?

$\displaystyle \log_n 1 = x$

$\displaystyle \Rightarrow n^x = 1$

Anything raised to the zero gives one, so the power is zero
• Jun 23rd 2007, 08:15 PM
Soroban
Hello, kevinz!

Quote:

Solve the following logarithmic equation:

$\displaystyle a)\;\log_2(6x-8) - \log_2(2x+4) \:=\:\log_2 (x-6)$

Here's a trick I've taught all my students . . .

If we have a logarithmic equation and a log-term is negative,
. . "move" it to the other side. .**

So we have: .$\displaystyle \log_2(6x-8) \;=\;\log_2(x-6) + \log_2(2x+4) \;=\;\log_2\left[(x-6)(2x+4)\right]$

. . . . .Then: .$\displaystyle \log_2(6x-8) \;=\;\log_2(2x^2-8x-24)$

Exponentiate (or "un-log") both sides: .$\displaystyle 6x - 8 \;=\;2x^2 - 8x - 24$

We have a quadratic: . . $\displaystyle x^2 - 7x - 8 \;=\;0$

. . . . . which factors: .$\displaystyle (x - 8)(x + 1) \;=\;0$

. . . . .and has roots: . . . . $\displaystyle x \:=\:8,\:(-1)$ .
-1 is extraneous

Therefore, the only root is: .$\displaystyle x \,=\,8$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** .Why? .To avoid that annoying quotient.

• Jun 23rd 2007, 08:28 PM
Jhevon
Quote:

Originally Posted by Soroban
Hello, kevinz!

Here's a trick I've taught all my students . . .

If we have a logarithmic equation and a log-term is negative,
. . "move" it to the other side. .**

So we have: .$\displaystyle \log_2(6x-8) \;=\;\log_2(x-6) + \log_2(2x+4) \;=\;\log_2\left[(x-6)(2x+4)\right]$

. . . . .Then: .$\displaystyle \log_2(6x-8) \;=\;\log_2(2x^2-8x-24)$

Exponentiate (or "un-log") both sides: .$\displaystyle 6x - 8 \;=\;2x^2 - 8x - 24$

We have a quadratic: . . $\displaystyle x^2 - 7x - 8 \;=\;0$

. . . . . which factors: .$\displaystyle (x - 8)(x + 1) \;=\;0$

. . . . .and has roots: . . . . $\displaystyle x \:=\:8,\:(-1)$ .
-1 is extraneous

Therefore, the only root is: .$\displaystyle x \,=\,8$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** .Why? .To avoid that annoying quotient.

Quotients are annoying, aren't they?! didn't seem like too big a deal in this case though, these were simple enough