# Thread: Need Help with Basic Log

1. ## Need Help with Basic Log

Evaluate:
1) $\displaystyle \log_2 128 - \log_2 32$
2) $\displaystyle \log_8 \sqrt8$

Can someone please help me with these log equations! Please show me what to do IN DETAIL (STEP BY STEP)! I really don't get what to do. I know you're suppose to do error and trial for the first equation, but how do I do this? And I don't know how to use the calculator "log" function when the base is not 10!

1) $\displaystyle 2$
2) $\displaystyle 0$

P.S.
Are the following the correct way of writing an inverse of $\displaystyle y = 4^x$?
Exponential:
$\displaystyle f^-1(x) = 4^x$
Logarithmic:
$\displaystyle y = \log_4 x$

2. Originally Posted by Macleef
Evaluate:
1) $\displaystyle \log_2 128 - \log_2 32$
2) $\displaystyle \log_8 \sqrt8$

Can someone please help me with these log equations! Please show me what to do IN DETAIL (STEP BY STEP)! I really don't get what to do. I know you're suppose to do error and trial for the first equation, but how do I do this? And I don't know how to use the calculator "log" function when the base is not 10!

1) $\displaystyle 2$
2) $\displaystyle 0$

P.S.
Are the following the correct way of writing an inverse of $\displaystyle y = 4^x$?
Exponential:
$\displaystyle f^-1(x) = 4^x$
Logarithmic:
$\displaystyle y = \log_4 x$
When trying to simplify basic logs of the form $\displaystyle \log_a b$, try to turn it into the form $\displaystyle \log_a a^x$, as this equals a.

1) $\displaystyle \log_2 128 -\log_2 32$

$\displaystyle =\log_2 2^7 - \log_2 2^5$

Now, think of what this means.

$\displaystyle x=\log_2 2^7 \iff 2^x = 2^7$

x must then be 7.

Also, $\displaystyle x=\log_2 2^5 \iff 2^x = 2^5$

$\displaystyle \Rightarrow x = 5$

So we then have $\displaystyle 7-5=2$

2) $\displaystyle \log_8 \sqrt{8}$

Recall that $\displaystyle \sqrt{8} = 8^\frac{1}{2}$

So, $\displaystyle \log_8 \sqrt{8} = \log_8 8^{\frac{1}{2}} = \frac{1}{2}$ ... i think the answers must be wrong.

3. Thanks, it makes sense to me now