SOLUTION: solve: log(base b) (x^2+7) = 2/3 log (base b) 64

Question 414944: solve: log(base b) (x^2+7) = 2/3 log (base b) 64
Found 2 solutions by Gogonati, jsmallt9:
Answer by Gogonati(855) (Show Source): You can put this solution on YOUR website!
log(x^2+7)=log2^6.2/3
x^2+7=4
x^2=-3 This equation doesn't have solution on R
The solution on complex set will be:x=ixsquare root of 3
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
oe
log(expression) = (other-expression)

All we have to do with your equation to achieve the second form is to get rid of the 2/3 somehow. Fortunately a property of logarithms, , allows us to move a coefficient of a logarithm into the argument as its exponent. Using this property we can mive the 2/3 into the argument:

We now have the second form. With the next step is based on some simple logic. The equation says that twp base b logarithms are equal. The only way two base b logarithms can be equal is if the arguments are equal, too. So:

You might already know how the right side simplifies. But if you have trouble with negative and/or fractional exponents, I find it can be helpful to factor the exponent in a certain way:
If the exponent is negative, factor out -1.
If the exponent is fractional and the numerator of the fraction is not a 1, then factor out the numerator. (You'll see this in a moment.)

Let's use this on
1) The exponent is not negative.
2) The exponent is fractional and its numerator is 2. So factor out 2:

With the exponent factored, each factor tells us an operation to perform<
The 2 as an exponent tells us to square (IOW: Raise to the 2nd power).
The 1/3 as an exponent tells us to find a cube root.

And since multiplcation is Commutative, we can do these two operations in an y order we choose. I choose to find the cube root first because:
so finding its cube root will be easy,
Squaring 64 is not appealing,

So
Now our equation is:

This is a quadratic equation so we want one side to be zero. Subtracting 16 from each side we get:

Now we factor (or use the Quadratic Formula). This factors easily since it is a difference of squares:
(x+3)(x-3) = 0
From the Zero Product Property we know that one of these factors must be zero. So:
x+3 = 0 or x-3 = 0
Solving these we get:
x = -3 or x = 3
Lastly we must check our answers. You must ensure that each solution makes all arguments of all logarithms positive. If a "solution" makes any argument zero or negative, then it must be rejected. These rejected solutions can occur any time and they do not mean a mistake was made! So even expert mathmeticians must their answers on these problems.

When checking, use the original equation:

Checking x = -3:

which simplifies to:

At this point we can see that both arguments are going to be positive. So there is no reason to reject this solution. This is the required part of the check. This rest of the check is optional and it will tell us if we made a mistake. You are welcome to finish the check.

Checking x = 3:

which simplifies to:

At this point we can see that both arguments are going to be positive. So there is no reason to reject this solution. Again, you are welcome to finish the check.

So the solution to your equation is
x = -3 or x = 3
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