Find the value oflog2^1/2+1(3-2^1/2)

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- Sep 2nd 2006, 08:59 PMJoyceLogarithms
Find the value of

**log**2^1/2+1(3-2^1/2) - Sep 2nd 2006, 10:03 PMCaptainBlackQuote:

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 - Sep 2nd 2006, 10:07 PMSoroban
Hello,Joyce!

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

Quote:

Find the value of**log**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 PMshubhadeep
I think he meant

$\displaystyle

\log_{\sqrt{2}+1}(3-\sqrt{2})

$ - Sep 2nd 2006, 10:38 PMCaptainBlackQuote:

Originally Posted by**shubhadeep**

+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 AMSoroban
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 $\displaystyle \log_{1.586}(2.414)$