+140 i need help with a math question but it is for a send in assingment so if at all possible could you make a similar equation to the one i give so it isn't cheating?
solve for x algebraically
2log2(x-6) - log2x = 3
2 log(x - 6) - log(x) = log((x - 6)^2) + log(1/x) = log((x - 6)^2/x):
log((x - 6)^2/x) = 3 log(2)
3 log(2) = log(2^3) = log(8):
log((x - 6)^2/x) = log(8)
Cancel logarithms by taking exp of both sides:
(x - 6)^2/x = 8
Multiply both sides by x:
(x - 6)^2 = 8 x
Subtract 8 x from both sides:
(x - 6)^2 - 8 x = 0
Expand out terms of the left hand side:
x^2 - 20 x + 36 = 0
The left hand side factors into a product with two terms:
(x - 18) (x - 2) = 0
Split into two equations:
x - 18 = 0 or x - 2 = 0
Add 18 to both sides:
x = 18 or x - 2 = 0
Add 2 to both sides:
x = 18 or x = 2
(2 log(x - 6))/(log(2)) - (log(x))/(log(2)) ⇒ (2 log(2 - 6))/(log(2)) - (log(2))/(log(2)) = ((0 + 2 i) π + 3 log(2))/(log(2)) ≈ 3. + 9.06472 i:
So this solution is incorrect
(2 log(x - 6))/(log(2)) - (log(x))/(log(2)) ⇒ (2 log(18 - 6))/(log(2)) - (log(18))/(log(2)) = 3:
So this solution is correct
The solution is:
Answer: |x = 18
+9673 Example:
\(2\log_2(x+1) - \log_2x = 2\\ \log_2(x+1)^2 -\log_2x=2\\ \log_2\left(\dfrac{(x+1)^2}{x}\right) = 2\\ \dfrac{(x+1)^2}{x} = 4\\ x^2 + 2x + 1 = 4x\\ x^2 - 3x + 1 = 0\\ x = \dfrac{-(-3)\pm \sqrt{(-3)^2-(4)(1)(1)}}{2(1)}=\dfrac{3\pm \sqrt5}{2}\\ x = \dfrac{3+\sqrt5}{2} \text{ or }\dfrac{3-\sqrt5}{2}\text{(rejected)}\\ x = \dfrac{3+\sqrt5}{2}\)
Try to do your own question now!!
