# Math Help - Need a few logarithm formulas...

1. ## Need a few logarithm formulas...

Or at least I think I do...

These are the two problems I have:

#1: $4^{4x}=2000; x=?$

#2: $4log(x-4)=16; x=?$

...so I basically just need to know what $x$ equals, and just plugging in numbers isn't very practical Can anyone help?

2. $4^{4x}=2000$ take the log of both sides
$log(4^{4x})=log(2000)$ but now, one of the log rules says that $log(a^b)=b*log(a)$

so $4x*log4=log2000$ and $4x=\frac{log2000}{log4}$ so just divide by 4 and x= $\frac{log2000}{4log4}$

in general though, people seem to like to reduce logs for no reason that I can see... so log4= $log2^2=2log2$ so x= $\frac{log2000}{8log2}$

$4log(x-4)=16$ divide by 4 and log(x-4)=4, now raise each side to a power of 10, so $10^{log(x-4)}=10^4$

so x-4=10000, and x=10004

3. Thanks man, I understand the second one, but for the first one, is x= $\frac{log2000}{8log2}$ the answer? Because when I put that into a calculator, I get $.1242136307$, and then when I add that value into the equation

$4^{4x}=2000$ as $4^{4 * .1242136307}$

I get $1.991297872$, which isn't close to 2,000...

4. $x=\frac{log2000}{8log2}$ when you put it in your calculator, you didn't use order of operations correctly so your calculator did this: $\frac{log2000}{8}*log2$

5. Ahhhh ok. I went back and entered it in as:
Code:
(log(2000))/(8log(2))
And got $1.370723036$, which worked perfectly...thanks for all of your help