1. ## quick log question

I was thinking about different log laws and stuff but i don't know how to get to that. It was on a past paper im pratising on at the moment.

If anyone can help, il really appreciate it.

2. ## Logs

Hello chinkmeista
Originally Posted by chinkmeista

I was thinking about different log laws and stuff but i don't know how to get to that. It was on a past paper im pratising on at the moment.

If anyone can help, il really appreciate it.

$\displaystyle \log_ax^{10} - 2\log_a\left(\frac{x^3}{4}\right)$

$\displaystyle = \log_ax^{10} -\log_a\left(\frac{x^3}{4}\right)^2$

$\displaystyle = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)$

$\displaystyle = \log_a\left(x^{10}\times\frac{16}{x^6}\right)$

$\displaystyle = \log_a(16x^4)$

$\displaystyle =4\log(2x)$

Hello chinkmeista
$\displaystyle \log_ax^{10} - 2\log_a\left(\frac{x^3}{4}\right)$

$\displaystyle = \log_ax^{10} -\log_a\left(\frac{x^3}{4}\right)^2$

$\displaystyle = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)$

$\displaystyle = \log_a\left(x^{10}\times\frac{16}{x^6}\right)$

$\displaystyle = \log_a(16x^4)$

$\displaystyle =4\log(2x)$

Oh hi, thats fantastic thank you. I kinda understand it, up until like the 3rd step though from:

$\displaystyle = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)$

to

$\displaystyle = \log_a\left(x^{10}\times\frac{16}{x^6}\right)$

how did you flip the 16 and x^6 upside down and merge the x^10 together at the same time? could you explain how you did this bit?

Again, thanks a lot.

4. Originally Posted by chinkmeista
Oh hi, thats fantastic thank you. I kinda understand it, up until like the 3rd step though from:

$\displaystyle = \log_ax^{10} -\log_a\left(\frac{x^6}{16}\right)$

to

$\displaystyle = \log_a\left(x^{10}\times\frac{16}{x^6}\right)$

how did you flip the 16 and x^6 upside down and merge the x^10 together at the same time? could you explain how you did this bit?

Again, thanks a lot.
$\displaystyle -\log_x\left(\frac{y}{z}\right) = \log\left(\left(\frac{y}{z}\right)^{-1}\right) = a\log\left(\frac{z}{y}\right)$

You can take the coefficient outside the log and put it into it's power. Anything to the power of $\displaystyle -1$ is the recipricole.

5. Originally Posted by chinkmeista
Oh hi, thats fantastic thank you. [snip]
No, no .... when I reply it's fantastic. When Grandad replies it's grand ......

6. ## Logs

Hello chinkmeista
Originally Posted by chinkmeista
how did you flip the 16 and x^6 upside down and merge the x^10 together at the same time? could you explain how you did this bit?[/COLOR][/COLOR][/COLOR]

Again, thanks a lot.
$\displaystyle \log a - \log b = \log (a \div b)$, and to divide by a fraction, turn it 'upside down' and multiply.

So $\displaystyle \log_a x - \log_a \left(\frac{y}{z} \right) = \log \left(x \times \frac{z}{y} \right )$.

Can you see it OK now?

7. ## Ho-ho

Originally Posted by mr fantastic
No, no .... when I reply it's fantastic. When Grandad replies it's grand ......
Ho! Ho!

$\displaystyle \log a - \log b = \log (a \div b)$, and to divide by a fraction, turn it 'upside down' and multiply.
So $\displaystyle \log_a x - \log_a \left(\frac{y}{z} \right) = \log \left(x \times \frac{z}{y} \right )$.