Can't I solve these using my calculator?

If not, can you help me get started? I'm drawing a total blank. Thanks!! :eek:

4^x=1/256

and

10^log(3x)=4

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- Nov 18th 2007, 12:24 PMcrazydaizy78logarithms/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!! :eek:

4^x=1/256

and

10^log(3x)=4 - Nov 18th 2007, 01:28 PMtopsquark
$\displaystyle 4^x = \frac{1}{256}$

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 - Nov 18th 2007, 01:30 PMtopsquark