SOLUTION: Simplify 0.001^(log2)...... what i think i need to do is just punch in the calculator log2 and then solve it, but my teacher wants to see me show my work. I hope you can help me,

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Simplify 0.001^(log2)...... what i think i need to do is just punch in the calculator log2 and then solve it, but my teacher wants to see me show my work. I hope you can help me,       Log On


   

Question 79927: Simplify 0.001^(log2)...... what i think i need to do is just punch in the calculator log2 and then solve it, but my teacher wants to see me show my work. I hope you can help me, and lastly thank you!
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
One way to look at it ... let's set the given problem equal to y and then solve for y.
.
So let's say:
.
y+=+0.001%5E%28log%282%29%29
.
But we can also say that 0.001 is equal to 10%5E%28-3%29. So substitute for 0.001 to get:
.
y+=+%2810%5E%28-3%29%29%5Elog%282%29
.
Now let's take the log of both sides to get:
.
log%28y%29+=+log%28%2810%5E%28-3%29%29%5Elog%282%29%29
.
But by the rules of logs the exponent (log(2)) can be written as the multiplier of the
log to give:
.
log%28y%29+=+log%282%29%2Alog%2810%5E%28-3%29%29
.
And again by the rules of logs, the exponent of the (-3) can be written
as the
multiplier of its log. This makes the equation become:
.
log%28y%29+=+log%282%29%2A%28%28-3%29%2Alog%2810%29%29
.
But recall that log(10) is equal to 1. So we can just replace log(10) by 1 in the first
term on the right side. This makes the equation become:
.
log%28y%29+=+log%282%29%2A%28-3%29%2A1
.
And by multiplying -3 * 1 it further simplifies to:
.
log%28y%29+=+log%282%29%2A%28-3%29
.
and rearranging the right side we get:
.
log%28y%29+=+-3%2Alog%282%29
.
Now in reverse of what we did previously regarding a rule of logs and exponents, we can
make -3 the exponent of the log to get:
.
log%28y%29+=+log%282%5E%28-3%29%29
.
Now let's look at 2%5E%28-3%29. Recall that a negative exponent means that the the term
should place in the denominator with a positive exponent. This means that 2%5E%28-3%29
is equivalent to 1%2F2%5E%283%29 and the denominator %282%5E3%29=8 so that 2%5E%28-3%29+=+1%2F8
.
Substituting this we get:
.
log%28y%29+=+log%281%2F8%29
.
Look carefully at this. For this equation to be true, y must be equal to 1%2F8
so that log%281%2F8%29+=+log%281%2F8%29
.
But going all the way back to the start ... we set out to find y ... and we now know that
y is equal to 1%2F8 and this is equivalent to 0.125
.
Now you can use your calculator to check the answer.
.
All that being said, I agree with you that it would have been easier to use a calculator
to work out the answer. However, the teacher is doing right in having you learn the
basics so that you really understand what the calculator is doing for you. If you just
rely on a calculator, you could easily fall into the trap of not understanding logs.