Question 272476: log base 2 multiply (x-2)+log base 2 multiply (8-x)-log base 2 multiply (x-5)=3
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
I'm guessing you are having trouble with logarithms. I say this because there are NO multiplications in your equation! It takes time to get used to functions. The basic form of functional notation is:
function-name(argument)
Before functions, whenever we saw two expressions without an operation symbol (+, -, *, /) between them we were supposed to assume the operation was multiplication. But once you start working with functions (log functions, trig functions (sin, cos, tan, etc.) and generic functions (f(x), g(x), etc.) you need to learn to identify function notation and understand that with function notation there is NO implied multiplication between the function name and its argument.
So the way to read/write your equation is:
"The base 2 log of x-2 plus the base 2 log of 8-x minus the base 2 log of x-5 is equal to 3". (And these of's do not mean multiplication like they often do in word problems.)
Back to your problem. With logarithmic equations that have variables in the arguments of logarithms, you want to transform your equation into one of the following forms:
log(variable-expression) = other-expression
or
log(variable-expression) = log(other-expression)
Since your equation has no logarithms on the right side, I am going to shoot for the first form. This form requires that there is only one logarithm on the left side and you have three logarithms. Somehow we need to combine the three logs into one. They are not like terms so we cannot simply add the first two and subtract the third. But fortunately we have some properties of logarithms which allow us to to do what we need:
The first property allows us to combine to logs with a "+" between them and the second one allows us to combine two logs with a "-" between them. We'll use the first property to combine your first two logarithms:
which simplifies to:
Now we can use the second property to combine the remaining logarithms:
We now have achieved the first form. With this form we proceed by rewriting the equation in exponential form:
which simplifies to:
Our logarithms are now all gone. We'll get rid of the fraction by multiplying both sides by x-5:
which simplifies to:
This is a quadratic equation so we want one side to be zero. Subtracting 8x from each side and adding 40 to each side we get:
Since we want to factor and since factoring a trinomial is easiest when the squared term has a coefficient of 1, I'm going to multiply both sides by -1:
Now we can factor this easily:
From the Zero Product Property we know that this product is zero only if one of the factors is zero. So:
x+4 = 0 or x-6 = 0
Solving these we get:
x = -4 or x = 6
With logarithmic equations it is important (not just a good idea) to check your answers. We must make sure that the solutions make all the arguments of logarithms positive. Always use the original equation to check.
Checking x = -4:
which simplifies to
As we can see, two of the arguments are negative. We cannot allow this so we must reject x = -4 as a solution.
Checking x = 6:
which simplifies to
None of the arguments are zero or negative so we are OK so far. These logs can be found by hand. Since   . And  by definition and since any number (except 0) to the zero power is 1,  for ALL bases. Substituting these in we get:
 Check!
So the only solution to your equation is x = 6.
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