# log equations

• May 12th 2008, 01:00 PM
nugiboy
log equations
I just did a past paper, and heres the two questions i couldn't do.

http://i137.photobucket.com/albums/q220/e3tiger/2.jpg

For the part 'i' question, i think its something to do with the unit circle, but im not sure. This is a non-calculator paper so i have to find it some other way. Or it might just be a question of learning it off by heart?

For the part 'ii' question, i can't seem to get it to work. I think i know the rules of logs pretty well. I started by dividing both terms by eachother as there is a - sign between them. Could someone give me help on that.

• May 12th 2008, 01:39 PM
Moo
Quote:

Originally Posted by nugiboy
I just did a past paper, and heres the two questions i couldn't do.

http://i137.photobucket.com/albums/q220/e3tiger/2.jpg

For the part 'i' question, i think its something to do with the unit circle, but im not sure. This is a non-calculator paper so i have to find it some other way. Or it might just be a question of learning it off by heart?

Because $\tan 300=\tan (360-60)$
Two possibilities :
- you transform tan into sin/cos :
you use the unit circle
you use the formulaes cos(a-b) and sin(a-b)
- you know the formula tan(a-b)

Quote:

For the part 'ii' question, i can't seem to get it to work. I think i know the rules of logs pretty well. I started by dividing both terms by eachother as there is a - sign between them. Could someone give me help on that.

We'll take care of the left hand side of the equation.
First of all, put every constant inside the logarithm.

\begin{aligned} 2 \log_a \frac{x^3}{4} &=\log_a \left[\left(\frac{x^3}{4}\right)^2\right] \\
&=\log_a \frac{x^6}{2^4} \end{aligned}

Now you can divide :

\begin{aligned} \log_a x^{10}-\log_a \frac{x^6}{2^4} &=\log_a \left(x^{10} \cdot \frac{2^4}{x^6} \right) \\
&=\log_a (2^4 \cdot x^4) \\
&=\dots \end{aligned}

(Sun)
• May 12th 2008, 01:51 PM
Soroban
Hello, nugiboy!

Quote:

2) Show that: . $\log_a\!\left(x^{10}\right) - 2\log_a\!\left(\frac{x^3}{4}\right) \;=\;4\log_a(2x)$

The left side is: . $\log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^3}{4}\right)^2 \;\;=\; \;\log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^6}{16}\right)$

. . $= \;\;\log_a\!\left(\frac{x^{10}}{\frac{x^6}{16}}\ri ght) \;\;=\;\;\log_a\!\left(16x^4\right) \;\;=\;\;\log_a(2x)^4 \;\;=\;\;4\log_a(2x)$

• May 12th 2008, 02:01 PM
nugiboy
Quote:

Originally Posted by Soroban
Hello, nugiboy!

The left side is: . $\log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^3}{4}\right)^2 \;\;=\; \;\log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^6}{16}\right)$

. . $= \;\;\log_a\!\left(\frac{x^{10}}{\frac{x^6}{16}}\ri ght) \;\;=\;\;\log_a\!\left(16x^4\right) \;\;=\;\;\log_a(2x)^4 \;\;=\;\;4\log_a(2x)$

Cheers man

Thats exactly what i did, but got stuck near the end. I got up to

$
\;\;\log_a\!\left(16x^4\right) \;\;$
$
$

, but im not sure how that gets to the next bit.
• May 12th 2008, 02:09 PM
Moo
Quote:

Originally Posted by nugiboy
Cheers man

Thats exactly what i did, but got stuck near the end. I got up to

$
\;\;\log_a\!\left(16x^4\right) \;\;$
$
$

, but im not sure how that gets to the next bit.

$16=2^4$

-> $16x^4=(2x)^4$

:p