# Logarithm questions

• June 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
• June 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)

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

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

Now we can just drop the logs:

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

I think you can take it from here

Quote:

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

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

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

...and i think you can solve that equation

Quote:

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

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

$\log_n 1 = x$

$\Rightarrow n^x = 1$

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

Quote:

Solve the following logarithmic equation:

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

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

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

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

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

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

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

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

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

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

• June 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: . $\log_2(6x-8) \;=\;\log_2(x-6) + \log_2(2x+4) \;=\;\log_2\left[(x-6)(2x+4)\right]$

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

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

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

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

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

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

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

** .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