SOLUTION: Hello I have a problem with logarithem equasions that have a variable (the variable is ussually squared) in the base. This one I have a particular problem with: {{{ log(x,7)+log(

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hello I have a problem with logarithem equasions that have a variable (the variable is ussually squared) in the base. This one I have a particular problem with: {{{ log(x,7)+log(      Log On


   

Question 716184: Hello I have a problem with logarithem equasions that have a variable (the variable is ussually squared) in the base.
This one I have a particular problem with:
+log%28x%2C7%29%2Blog%28x%5E2%2C7%29=6+
Thank you for your time
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
+log%28x%2C%287%29%29%2Blog%28x%5E2%2C%287%29%29=6+
Most of the time the only way to manipulate multiple logarithms is when they have the same bases. So that is where we will start. We will try to get the bases of the logarithms to match.

For this we will be using the change of base formula: log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29. We can either change the first log so its base is the same as the second one or vice versa. I'm going to change the second log's base from x%5E2 to x:
+log%28x%2C%287%29%29%2Blog%28x%2C%287%29%29%2Flog%28x%2C+%28x%5E2%29%29=6+
We can find the numeric value of the log in the denominator by hand. That log represents the exponent one would put on x to get x%5E2. Obviously we must put an exponent of 2 on x to get x%5E2:
+log%28x%2C%287%29%29%2Blog%28x%2C%287%29%29%2F2=6+
To eliminate the fraction I'll multiply both sides of the square by 2:
+2log%28x%2C%287%29%29%2Blog%28x%2C%287%29%29=12+
The two logs on the left are like terms. (Like logarithmic terms have the same bases and the same arguments.) So we can add them together. Exactly like 2y + y = 3y our logs add up to:
+3log%28x%2C%287%29%29=12+
Dividing by 3 we get:
+log%28x%2C%287%29%29=4+

Once a logarithmic equation is in the form:
log(expression) = number
the next step is to rewrite the equation in exponential form. In general log%28a%2C%28p%29%29+=+n is equivalent to p+=+a%5En. Using this pattern on our equation we get:
7+=+x%5E4
Last we find the 4th root of each side. (Normally when finding a 4th (or any even) root, you have to remember both the positive and negative roots. But in this case we can ignore the negative root. In the original equation x is the base of the first log. Bases of logs cannot be negative. This is why we get to ignore the negative root.) So our solution is:
root%284%2C+7%29+=+x