SOLUTION: What is the value of logbase2 (1/sqrt32?

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Question 127474: What is the value of logbase2 (1/sqrt32?
Found 2 solutions by josmiceli, bucky:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C1%2Fsqrt%2832%29%29
writing this another way,
2%5Ex+=+1%2Fsqrt%2832%29
2%5Ex+=+1%2Fsqrt%2816%2A2%29
2%5Ex+=+1%2F%284%2A%282%5E%281%2F2%29%29%29
2%5Ex+=+2%5E%28-2%29%2A2%5E%28-1%2F2%29
2%5Ex+=+2%5E%28-2-%281%2F2%29%29
2%5Ex+=+2%5E%28-%285%2F2%29%29
x+=+-%285%2F2%29
log%282%2C1%2Fsqrt%2832%29%29+=+-%285%2F2%29 answer

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Set up the equation:
.
y+=+log%282%2C%281%2F%28sqrt%2832%29%29%29%29
.
Convert 1%2Fsqrt%2832%29 to a number by using a calculator to divide the square root of 32
into 1. The answer to this division is 0.176776695. Substitute this into the equation
to get:
.
y+=+log%282%2C0.176776695%29
.
Get this into a little more conventional form by transposing (or switching) sides to get:
.
+log%282%2C0.176776695%29+=+y
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Now convert this logarithmic form to its equivalent form by raising the base of the logarithm
(the base being 2) to the y power and setting this equal to 0.176776695. In other words
you are converting the above logarithmic equation to the form:
.
2%5Ey+=+0.176776695
.
Then take the logarithm of both sides ... only this time let's use base 10 as the logarithm
because we can work in base 10 on a calculator. When you take the logarithm base 10 of both
sides the equation becomes:
.
log%2810%2C2%5Ey%29+=+log%2810%2C0.176776695%29
.
On the left side of this equation, use the exponent rule of logarithms to bring out the
exponent y as a multiplier of the logarithm. This results in the equation becoming:
.
y%2Alog%2810%2C2%29+=+log%2810%2C0.176776695%29
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Now use your calculator to find that log%2810%2C0.176776695%29+=+-0.752574989 and also
to find that log%2810%2C2%29+=+0.301029995.
.
Substitute these numbers into the equation in place of their logarithm forms and you have:
.
y%2A0.301029995+=+-0.752574989
.
Solve for y by dividing both sides of this equation by 0.301029995 and the result is:
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y+=+-0.752574989%2F0.301029995+=+-2.5
.
And recall way back at the start of this exercise we said that y was equal to log%282%2C%281%2Fsqrt%2832%29%29%29.
.
So we can now say that log%282%2C%281%2Fsqrt%2832%29%29%29=+-2.5
.
and the answer to this problem is -2.5
.
Hope this helps you to understand the problem and one way of doing it. The process that
was used above to solve the problem essentially was the same process you use to develop the
change of base formula that you might be familiar with.
.