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For most problems it helps to simplify the expressions before you start trying to solve the equation. If you understand what logarithms are you will see that will simplify. What is and the equation will simplify when we replace this log with 2:
This is all the simplifying that can be done so we are ready to start solving the equation.
In general, to solve equations where the variable is in the argument of one or more logarithms, you transform the equation into one of the following forms:
log(expression-with-variable) = another-expression
or
log(expression-with-variable) = log(another-expression)
Since there is a term that is not a logarithm, the 3, we will aim for the first form. To get your equation into this form we need to get all the logarithms combined into one. Fortunately we have properties which allow us to combine logarithms:
These properties require that there be no (visible) coefficient. (In other words the coefficient of the logarithm must be 1.) And fortunately there is a property we can use to handle unwanted coefficients: .
So we will gather all the logarithms to one side of the equation and then use these properties to combine them into one. First we will subtract and 4 from each side:
Next we will use the property involving the subtraction of logarithms to combine the logs:
Now we have the desired form. Next we reqrite the equation in exponential form. To do this we have to remember the the connection between logarithmic equations and exponential equations: is equivalent to . Using this on our equation we get:
Since we have:
Now we have an equation, without logarithms, that we can solve. One way to do this is to multiply both sides by :
By doing this you can see that both denominators will cancel out:
leaving:
which simplifies to:
Adding 6y to each side we get:
Subtracting 1 from each side we get:
Dividing by 5 we get:
With logarithmic equations it is more than just a good idea to check your answers. It is important. You need to make sure that the solution(s) do not make the arguments of any logarithms zero or negative.
Checking y = 3/5:
At this point we can see that the arguments of both logarithms are positive. So there is no reason to reject our solution. If you want to finish the check to see that 3/5 actually works, you will have to figure out with this equation is actually true. To do this we can either use the change of base formulas to change these base 2 logs into a base your calculator can do (usually 10 and/or e) or we can be a little clever:
Factor a 2 in the argument of the log on the right:
Use one the properties to split the log on the right into two logs:
Since log(2, (2)) = 1 we have:
Adding the 3 and the 1 we get: Check!!
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