SOLUTION: What is the answer to log5+2logx=log45 when trying to solve for x

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Question 607582: What is the answer to log5+2logx=log45 when trying to solve for x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%285%29%29%2B2log%28%28x%29%29=log%28%2845%29%29
Solving equations where the variable is in the argument (or base) of a logarithm usually starts with using algebra and/or properties of logarithms to transform the equation into on of the following forms:
log(expression) = other_expression
or
log(expression) = log(other_expression)

IF we could combine the two terms on the left side into one, we would have the second form. They are not like terms so we cannot just add them. There is a property of logarithms, log%28a%2C+%28p%29%29+%2B+log%28A%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29, which offers another way to combine two logs which have a "+" between them. But this property requires that the log's have coefficients of 1. The second log on the left side has a coefficient of 2. Fortunately there is another property of logarithms, q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29, which allows us to move a coefficient of a log into the argument as its exponent. By using this property we can then use the other property to combine the terms:
log%28%285%29%29%2Blog%28%28x%5E2%29%29=log%28%2845%29%29
log%28%285%2Ax%5E2%29%29=log%28%2845%29%29
We now have the second form.

The next step with the second form is based on some simple logic: The only way two logarithms of the same base can be equal is if the arguments are equal. So:
5%2Ax%5E2=45
We now have an equation without logs and where the variable is "out in the open". Now we use algebra to solve for x. Since this is a quadratic equation, we want one side to be zero. Subtracting 45 from each side we get:
5%2Ax%5E2-45+=+0
Now we factor (or use the Quadratic Formula). First the GCF:
5%28x%5E2-9%29+=+0
Next we have a difference of squares:
5%28x%2B3%29%28x-3%29+=+0
From the Zero Product Property we know that one (or more) of these factors much be zero. So:
x + 3 = 0 or x - 3 = 0
Solving each of these we get:
x = -3 or x = 3

Checking solutions for equations like this one is not optional. You must ensure that all arguments to all logarithms are positive. Use the original equation to check:
log%28%285%29%29%2B2log%28%28x%29%29=log%28%2845%29%29
Checking x = -3:
log%28%285%29%29%2B2log%28%28%28-3%29%29%29=log%28%2845%29%29
We can already see that the argument to the second logarithm is be negative when x = -3. So we must reject this solution.
Checking x = 3:
log%28%285%29%29%2B2log%28%28%283%29%29%29=log%28%2845%29%29
We can already see that the argument to all three logarithms are positive when x = 3. So this is a solution. Since it is the only solution we did not reject, it is the only solution to your equation.