SOLUTION: How would I solve this? log2(x-3)-log2(2x+1)=log24 (log base 2, 4)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: How would I solve this? log2(x-3)-log2(2x+1)=log24 (log base 2, 4)      Log On


   

Question 812286: How would I solve this?
log2(x-3)-log2(2x+1)=log24 (log base 2, 4)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I'm guessing that the reason you have not received a reply so far is that it is very difficult to know what the equation is. Is it
log%282%2C+%28x-3%29%29-log%282%2C%282x%2B1%29%29=log%2824%2C+%28log%282%2C+%284%29%29%29%29
or is that 24 really supposed to be a 2?
log%282%2C+%28x-3%29%29-log%282%2C%282x%2B1%29%29=log%282%2C+%28log%282%2C+%284%29%29%29%29
or something else?

Logarithms can be difficult to enter in a clear way on algebra.com. One option is to use some English, like you did when you used "base 2". Another option is to learn algebra.com's syntax for displaying formulas. To see how I got algebra.com to display the logs above, click on the "Show source" link which should be just above this reply.

I'm going to take a chance that the second equation is the correct one. First I will simplify. Since 4 is 2 squared, the rightmost log is a 2:
log%282%2C+%28x-3%29%29-log%282%2C%282x%2B1%29%29=log%282%2C+%282%29%29
The rightmost log can now be simplified:
log%282%2C+%28x-3%29%29-log%282%2C%282x%2B1%29%29=1

Now we'll start solving.... In general, the first part of solving logarithmic equations is to use algebra and/or properties of logs to transform the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

If we could find a way to combine the two logs on the left side we would have the first form. Fortunately one of the properties of logs, log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29 which will let us combine the terms:
log%282%2C+%28%28x-3%29%2F%282x%2B1%29%29%29=1
And we now have the first form.

The next part of solving an equation in the first form is to rewrite the equation in exponential form. Since log%28a%2C+%28p%29%29+=+n is equivalent to p+=+a%5En our equation can be rewritten as:
%28x-3%29%2F%282x%2B1%29=2%5E1
which simplifies to:
%28x-3%29%2F%282x%2B1%29=2

Now that the logs are gone, we can now use "regular" algebra to solve. First we'll multiply each side by 2x + 1 in order to eliminate the fraction:
x-3=4x%2B2
Subtracting x:
-3+=+3x%2B2
Subtracting 2:
-5+=+3x
Dividing by 3:
%28-5%29%2F3+=+x

Now we check. This is not optional! A check must be made to see that all bases and arguments of all logarithms remain valid. (Valid bases are any positive numbers except 1 and valid arguments are any positive number.) Any "solution" that makes a base or an argument invalid must be rejected.

Using the original equation:
log%282%2C+%28x-3%29%29-log%282%2C%282x%2B1%29%29=log%282%2C+%28log%282%2C+%284%29%29%29%29
to check x+=+-5%2F3:

Simplifying...


We cal already see that both arguments on the left side are invalid. So we must reject this solution. (If only one argument had been invalid we would still reject the "solution".) Note: We are rejecting x = -5/3 because the arguments are invalid, not because x is negative.

Rejecting the only solution we had found means one of two things:
  • There are no solutions to the equation. Or...
  • My guess about the equation is wrong. If this is the case then perhaps the steps I've shown above will help you with the correct equation.