Log(x-6) +Log(x+3)=1
Solve for x
I think it has no solution but why?
well, work it out...
log(x-6)+log(x+3) = 1
log (x-6)*(x-3) = 1
(x-6)*(x-3) = 10
x^2-9x+18 = 10
x^2-9x+8 = 0
(x-1)(x-8) = 0
x = 1 or 8
x=1 does not work, because then x-6 < 0 and the log is not defined. So, the only real solution is x=8
Check:
log(2)+log(5) = 1
log(10) = 1
yes.
Hmmm. Let's try that again without the typo...
log (x-6)*(x+3) = 1
x^2-3x-18 = 10
x^2-3x-28 = 0
x = (3±11)/2
x = -4 or 7
The only real solution is x=7
Check:
log(1)+log(10) = 1
yes
To solve the given equation, we can apply the properties of logarithms.
First, we can use the property that states:
log(a) + log(b) = log(ab)
Applying this property to the equation, we have:
log((x-6)(x+3)) = 1
Next, we can use the inverse property of logarithms, where:
log(base a) a^b = b
Using this property, we can rewrite the equation as an exponent:
(x-6)(x+3) = 10
Expanding the left side of the equation, we get:
x^2 - 3x - 18 = 10
Rearranging the equation, we have:
x^2 - 3x - 28 = 0
Now we can solve this quadratic equation. We can factor it as:
(x - 7)(x + 4) = 0
Setting each factor equal to zero, we get:
x - 7 = 0 or x + 4 = 0
Solving each equation separately:
For x - 7 = 0, adding 7 to both sides gives:
x = 7
For x + 4 = 0, subtracting 4 from both sides gives:
x = -4
So we have two potential solution values: x = 7 and x = -4.
Now, we need to check if these solutions are valid in the original equation.
Since the given equation contains logarithms, we need to ensure that the arguments of the logarithms are positive.
For x = 7, substituting back into the equation:
log(7-6) + log(7+3) = 1
log(1) + log(10) = 1
0 + 1 = 1
1 = 1
The equation holds true, so x = 7 is a valid solution.
Now, let's check x = -4:
log(-4-6) + log(-4+3) = 1
log(-10) + log(-1) = 1
Here, we encounter a problem. Logarithms are only defined for positive arguments. Since the argument of the logarithm is negative, there is no valid solution for x = -4.
Hence, the original equation has only one solution, which is x = 7. The solution x = -4 is not valid.