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

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- September 2nd 2006, 09:59 PMJoyceLogarithms
Find the value of

**log**2^1/2+1(3-2^1/2) - September 2nd 2006, 11:03 PMCaptainBlackQuote:

Originally Posted by**Joyce**

This could mean:

,

or

or any number of other things

RonL - September 2nd 2006, 11:07 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:

If you see , do**not**write it as: log 2^8, which means

The "2" is a*subscript*, written 'below the line'.

. . You can write: log_2(8) . . . okay?

- September 2nd 2006, 11:15 PMshubhadeep
I think he meant

- September 2nd 2006, 11:38 PMCaptainBlackQuote:

Originally Posted by**shubhadeep**

+rep today :confused:

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

RonL - September 3rd 2006, 05:49 AMSoroban
I too like Shubadeep's interpretation.

But I*hope*that there are typos in the original problem.

If it was: . . . . the answer is .

Otherwise, I see no point to the problem

. . other than using the Base-Change Formula

. . which could have been practiced with