- #1
storoi1990
- 14
- 0
Q: Logx = 1 + 6/logx
my attempt so far: Log x^2 = 7
where do i go from here?
Is this [itex]log x= 1+ \frac{6}{log x}[/itex] or [itex]log x= \frac{1+ 6}{log x}[/itex]?storoi1990 said:Q: Logx = 1 + 6/logx
my attempt so far: Log x^2 = 7
where do i go from here?
You mean the first one, which is logx = 1 + (6/logx).storoi1990 said:yeah it's the last one.
storoi1990 said:so then i end up with:
y^2-y-6 = 0
[\quote]This factors into (y - 3)(y + 2) = 0, so y = 3 or y = -2.
Now, since y = logx, you have logx = 3 or log x = -2.
What are the solutions for x? They are NOT 3 and -2, as you show below.
storoi1990 said:x1 = 3 , and x2 = -2
is that the answer?
A logarithm is the inverse function of exponentiation. It is used to solve equations where the variable is in the exponent.
The base of a logarithm is the number that is raised to a certain power to get the argument of the logarithm. In this equation, the base is unknown and represented as logx.
To solve for log x^2 = 7, we need to use the properties of logarithms to rewrite the equation as x^2 = 10^7. Then, we can take the square root of both sides to get x = ±√10^7. This can be simplified to x = ±√10 * 10^3, which means x = ±√10 * 1000. Finally, we can simplify further to get x = ±10√10.
Since we took the square root of both sides, there are two possible solutions for x: x = 10√10 or x = -10√10.
Yes, this equation can be solved without a calculator by using the properties of logarithms and simplifying the equation algebraically. However, using a calculator can help to find the approximate decimal value of the solution.