Solve the equation:
Logx=1-log(x-3)
log x + log (x-3) = 1
log (x^2 -3x) = 1
10^log (x^2 -3x) = 10^1 = 10
x^2 -3 x -10 = 0
(x-5)(x+2) = 0
x = +5 or - 2
but -2 is not a solution, since the domain of log(x) is x>0.
To solve the equation log(x) = 1 - log(x - 3), we can use properties of logarithms and algebraic manipulation. Here's how you can solve it step by step:
Step 1: Apply the logarithmic rules
Using the logarithmic rules, we can rewrite the equation as follows:
log(x) + log(x - 3) = 1
Step 2: Combine the logarithms
By using the property log(a) + log(b) = log(ab), we obtain:
log[x(x - 3)] = 1
Step 3: Rewriting in exponential form
Applying the definition of logarithms, we can rewrite the equation in exponential form:
x(x - 3) = 10^1
Simplifying the right side:
x(x - 3) = 10
Step 4: Expand and rearrange
Expanding the left side of the equation:
x^2 - 3x = 10
Rearranging it into a quadratic equation:
x^2 - 3x - 10 = 0
Step 5: Solve the quadratic equation
Now, we need to solve the quadratic equation x^2 - 3x - 10 = 0. This can be done by factoring or applying the quadratic formula. Let's solve it using factoring.
The factors of -10 that add up to -3 are -5 and 2:
(x - 5)(x + 2) = 0
Setting each factor equal to zero and solving for x:
x - 5 = 0 => x = 5
x + 2 = 0 => x = -2
Step 6: Check for extraneous solutions
It is important to check if any of the solutions we obtained are extraneous, i.e., they do not satisfy the original equation.
Let's check both solutions:
For x = 5:
log(5) = 1 - log(5 - 3)
log(5) = 1 - log(2)
This equation is true, so x = 5 is a valid solution.
For x = -2:
log(-2) = 1 - log(-2 - 3)
The logarithm of a negative number is undefined, so x = -2 is not a valid solution.
Step 7: Final solution
Therefore, the solution to the equation log(x) = 1 - log(x - 3) is x = 5.