SOLUTION: solve for x: 〖log〗_2 x=〖log〗_4 (x^2-6)

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Question 387558: solve for x:
〖log〗_2 x=〖log〗_4 (x^2-6)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
〖log〗_2 x=〖log〗_4 (x^2-6)
This is very difficult to understand. You might get a quicker response to your posts if they were easy to understand. In particular, the 〖 and 〗 symbols are very confusing.

I am assuming that the equation is:
log%282%2C+%28x%29%29+=+log%284%2C+%28x%5E2-6%29%29
(You might want to "view source" to see how I entered this equation so it looks like it does,) This problem would be easy if the bases of the logarithms were the same. So we will start by using the base conversion formula, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, to convert one of the logarithms into the base of the other. I am going to convert the base 4 logarithm into an expression of base 2 logarithms:
log%282%2C+%28x%29%29+=+log%282%2C+%28x%5E2-6%29%29%2Flog%282%2C+%284%29%29
Since 4+=+2%5E2, log%282%2C+%284%29%29+=+2 so the equation becomes:
log%282%2C+%28x%29%29+=+log%282%2C+%28x%5E2-6%29%29%2F2
Multiplying both sides by 2 (to eliminate the fraction) we get:
2%2Alog%282%2C+%28x%29%29+=+log%282%2C+%28x%5E2-6%29%29
Next we can use a property of logarithms, q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29, to move the 2 from in front into the argument as an exponent:
log%282%2C+%28x%5E2%29%29+=+log%282%2C+%28x%5E2-6%29%29
Now we have an equation that says that one base 2 logarithm is equal to another. This can only be true if the arguments are equal. So:
x%5E2+=+x%5E2+-+6
Subtracting x%5E2 from each side we get:
0 = -6
The variable is gone and we have an "equation" which is a false statement. This means that there is no solution to this equation.