SOLUTION: Logx+1/2log16=log64

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Question 574315: Logx+1/2log16=log64
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%28x%29%29%2B%281%2F2%29log%28%2816%29%29=log%28%2864%29%29
Solving equations where the variable is in the argument (or base) of logarithm usually starts by transforming the equation into one of the following forms:
log(expression) = other_expression
or
log(expression) = log(other_expression)

Since your equation has logs on both sides of the equation, the second form looks easiest to reach. All we have to do is find a way to combine the two logs on the left side into one.

There are two properties of logarithms that can be used to combine logarithms:
  • log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29
  • log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29
These properties require that the coefficients of the the logs be 1's. Your second log has a coefficient of 1/2.

Fortunately there is another property of logarithms, q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+p%5Eq%29%29, which can be used to "move" a coefficient into the argument as its exponent. Using this property on the second log on the left side we get:
log%28%28x%29%29%2Blog%28%2816%5E%281%2F2%29%29%29=log%28%2864%29%29
Since an exponent of 1/2 means square root we can rewrite this as:
log%28%28x%29%29%2Blog%28%28sqrt%2816%29%29%29=log%28%2864%29%29
And since the square root of 16 is 4 this becomes:
log%28%28x%29%29%2Blog%28%284%29%29=log%28%2864%29%29

The left side is now in the form we need (with coefficients of 1's) for the "combine terms" properties. Since your equation has a "+" between the two terms we will use the first property to combine them:
log%28%28x%2A4%29%29=log%28%2864%29%29
or
log%28%284x%29%29=log%28%2864%29%29
We have now achieved the second form mentioned back at the beginning.

With the second form the next step is based on some simple logic: The equation says that two base 10 (aka "common") logs are equal. The only way this can be true is for the two arguments to be equal. So...
4x = 64

With the logarithms now gone this is a very simple equation to solve. Just divide both sides by 4:
x = 16

Last of all, check your answer. This is not optional when solving logarithmic equations! You must ensure that all arguments and bases are valid. Arguments and bases of logarithms can never be negative! (And the base can never be a 1, either.) Any solution, even if it is the only one you find, must be rejected if it makes an argument or base invalid!

Use the original equation to check:
log%28%28x%29%29%2B%281%2F2%29log%28%2816%29%29=log%28%2864%29%29
Checking x = 16:
log%28%2816%29%29%2B%281%2F2%29log%28%2816%29%29=log%28%2864%29%29
We can already see that all three arguments and bases are valid. So there is no reason to reject this solution. (You are welcome to use your calculator to complete the check. If the answer doesn't check it means we made a mistake that should be corrected.)