# Logarithms

• Sep 2nd 2006, 08:59 PM
Joyce
Logarithms
Find the value oflog 2^1/2+1(3-2^1/2)
• Sep 2nd 2006, 10:03 PM
CaptainBlack
Quote:

Originally Posted by Joyce
Find the value of log 2^1/2+1(3-2^1/2)

You need to use brackets to make this unambiguous.

This could mean:

$\displaystyle \log(2^{1/2}+1)\times (3-2^{1/2})$,

$\displaystyle \log[(2^{1/2}+1)\times (3-2^{1/2})]$

or

$\displaystyle \log((2^1/2+1)(3-2^{1/2}))$

or any number of other things

RonL
• Sep 2nd 2006, 10:07 PM
Soroban
Hello,Joyce!

I'd like to help, but I can't read the problem . . .

Quote:

Find the value oflog 2^1/2+1(3-2^1/2)

Did you even look at what showed up on the screen?

By the way, there should be no exponent in your log statement.

You probably meant something like: $\displaystyle \log_2\left(\frac{1}{2}\right)$

If you see $\displaystyle \log_28$, do not write it as: log 2^8, which means $\displaystyle \log 2^8$

The "2" is a subscript, written 'below the line'.
. . You can write: log_2(8) . . . okay?

• Sep 2nd 2006, 10:15 PM
I think he meant
$\displaystyle \log_{\sqrt{2}+1}(3-\sqrt{2})$
• Sep 2nd 2006, 10:38 PM
CaptainBlack
Quote:

I think he meant
$\displaystyle \log_{\sqrt{2}+1}(3-\sqrt{2})$

Good idea, I would give you +rep for that but I can't give you more
+rep today :confused:

Then by the change of logarithm base rule, this can be written:

$\displaystyle \log_{\sqrt{2}+1}(3-\sqrt{2})=\frac{\log_e(3-\sqrt{2})}{\log_e(\sqrt{2}+1)}$$\displaystyle \approx 0.5231$

RonL
• Sep 3rd 2006, 04:49 AM
Soroban
If it was: .$\displaystyle \log_{1+\sqrt{2}}(3 + 2\sqrt{2})$ . . . the answer is $\displaystyle 2$.
. . which could have been practiced with $\displaystyle \log_{1.586}(2.414)$