Find the value oflog 2^1/2+1(3-2^1/2)
You need to use brackets to make this unambiguous.Originally Posted by Joyce
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
Hello,Joyce!
I'd like to help, but I can't read the problem . . .
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?
Good idea, I would give you +rep for that but I can't give you moreOriginally Posted by shubhadeep
+rep today
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
I too like Shubadeep's interpretation.
But I hope that there are typos in the original problem.
If it was: .$\displaystyle \log_{1+\sqrt{2}}(3 + 2\sqrt{2}) $ . . . the answer is $\displaystyle 2$.
Otherwise, I see no point to the problem
. . other than using the Base-Change Formula
. . which could have been practiced with $\displaystyle \log_{1.586}(2.414)$