Where appropriate, include the approximation to the nearest ten-thousandth.
31.) log(x-4)+ log(x+4)= log4
I have the answer as 4
Is this answer right?
log a + log b is same as log ab
so log(x^2-4)=log4
taking antilogs
x^2-4=4 solve for 4
check my thinking.
Okay x^2-4 = 4
x^2= 8
sqrt(x^2)= sqrt 8
x= 2.83 is this right?
log k has a nasty habit of becoming undefined if k is not positive.
you answer of x = 2.832 is correct according to the steps taken, but...
when we sub into log (x-4) we get log (-1.168 which is undefined,
so there is no solution in the set of real numbers for your equation
Going back and looking over the previous solutions, I noticed that the original non-log equation should have said:
x^2-16 = 4
so x^2 = 20
x = √20 = 4.472136
which works upon substituting.
To determine if the answer is correct, we can solve the equation and check if it satisfies the given equation.
First, let's combine the logarithms using the logarithmic property that states log(a) + log(b) = log(a * b):
log((x - 4) * (x + 4)) = log(4)
Since the logarithms on both sides of the equation have the same base (log base 10 in this case), we can equate the arguments inside the logarithms:
(x - 4) * (x + 4) = 4
Expanding the left side of the equation:
(x^2 - 16) = 4
Adding 16 to both sides:
x^2 = 20
Taking the square root:
x = ±√20
However, we need to double-check if these solutions satisfy the given equation (log(x-4) + log(x+4) = log(4)).
If we substitute x = √20:
log((√20 - 4) + (√20 + 4)) = log(4)
log(16) = log(4)
The logarithms are not equal, so x = √20 does not satisfy the equation.
If we substitute x = -√20:
log((-√20 - 4) + (-√20 + 4)) = log(4)
log(-16) = log(4)
Logarithm of a negative number is undefined, so x = -√20 is not a valid solution.
Thus, both solutions do not satisfy the equation log(x-4) + log(x+4) = log(4).
In conclusion, the answer 4 is not the correct solution.