What is the sum of all possible real values of x that satisfies the equation
x
x^log5 =x^3/25.
What does that lone x just hanging around there ?
if you mean
x^log5 = x^(3/25)
then log 5 = 3/25 , which makes no sense
What is the sum of all possible real values of x that satisfies the equation xlog5x=x325.
Still not clear enough
is it
xlog(5x) or xlog5x ?
on the right side, originally you had x^(3/25
now I see
x325
is it x^325 or x^(3/25) like before ????
To find the sum of all possible real values of x that satisfy the equation, we need to solve the equation first.
Let's start by simplifying the equation. We have:
x^(log5) = x^(3/25)
Since the bases on both sides of the equation are the same, we can equate the exponents:
log5 = 3/25
To solve this equation, we need to isolate the variable. We can do that by taking the exponentiation of both sides with a base of 5:
5^(log5) = 5^(3/25)
On the left side, the base 5 and the exponent log5 cancel out, leaving us with:
5 = 5^(3/25)
Now, let's solve for x:
5^(3/25) is equivalent to taking the cube root of 5 raised to the power of 3:
5^(3/25) = (5^3)^(1/25) = 125^(1/25)
Now, we can write this equation as:
5 = 125^(1/25)
To get rid of the exponent on the right side, we raise both sides of the equation to the power of 25:
5^25 = (125^(1/25))^25
5^25 is equal to 9,765,625, and using the property of exponents, the right side simplifies to:
125^(1/25 * 25) = 125
So we have:
9,765,625 = 125
This equation is not true, which means there is no real value of x that satisfies the original equation.
Therefore, the sum of all possible real values of x is zero.