1. ## logarithms/calculator/solving

Can't I solve these using my calculator?
If not, can you help me get started? I'm drawing a total blank. Thanks!!

4^x=1/256

and

10^log(3x)=4

2. Originally Posted by crazydaizy78
Can't I solve these using my calculator?
If not, can you help me get started? I'm drawing a total blank. Thanks!!

4^x=1/256
$\displaystyle 4^x = \frac{1}{256}$

It helps if you know that $\displaystyle 256 = 4^4$. Then
$\displaystyle 4^x = \frac{1}{4^4} = 4^{-4}$

so x = -4.

To do this if you don't know that little fact:
$\displaystyle 4^x = \frac{1}{256}$

$\displaystyle log_4(4^x) = log_4 \left ( \frac{1}{256} \right )$ <-- If you use $\displaystyle log_{10}$ or ln here then you don't have to use the change of base formula a little further on.

$\displaystyle x = log_4(1) - log_4(256)$

$\displaystyle x = -log_4(256)$

Now let's say you still haven't seen that $\displaystyle 256 = 4^4$. Use the change of base formula to change the base to e (or 10 if you desire):
$\displaystyle log_a(b) = \frac{ln(a)}{ln(b)}$

So we have:
$\displaystyle x = -\frac{ln(256)}{ln(4)}$
which you can plug into your calculator and get that x = -4.

-Dan

3. Originally Posted by crazydaizy78
10^log(3x)=4
The ^ and log operators are inverses of each other, so in general
$\displaystyle a^{log_a(b)} = b = log_a(a^b)$

thus, taking "log" to be log base 10:
$\displaystyle 10^{log(3x)} = 4$

$\displaystyle 3x = 4$
which I'm sure you can solve from here.

-Dan