SOLUTION: If log sub2 x=7, evalutate log sub2 16x^2 So 2^7 then would it be 2^16x^2?? I'm so confused on even how to begin??

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: If log sub2 x=7, evalutate log sub2 16x^2 So 2^7 then would it be 2^16x^2?? I'm so confused on even how to begin??      Log On


   



Question 276066: If log sub2 x=7, evalutate log sub2 16x^2 So 2^7 then would it be 2^16x^2?? I'm so confused on even how to begin??
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
If log%282%2C+%28x%29%29=7, evaluate log%282%2C+%2816x%5E2%29%29.
What we want is to find a way to express the second logarithm in terms of the first. So somehoe we need to "peel away" the coefficient 16 and the exponent of 2. Fortunately we have properties of logarithms which allow us to do just that:
  • log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29 which allows us to split the log of a product into the sum of the logs of the factors.
  • log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29 which allows us to take the exponent of the argument and move it in front of the log.

log%282%2C+%2816x%5E2%29%29
Let's start by using the first property to split the product of 16 and x%5E2 into separate logas:
log%282%2C+%2816%29%29+%2B+log%282%2C+%28x%5E2%29%29
Since 2%5E4+=+16 the first log is 4:
4+%2B+log%282%2C+%28x%5E2%29%29
Now we can use the second property to move the exponent of x%5E2 out in front of the log:
4+%2B+2log%282%2C+%28x%29%29
At finally, since we are told that log%282%2C+%28x%29%29+=+7, we can substitute 7 for the for the remaining log:
4+%2B+2%2A7
which simplifies as follows:
4%2B14
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