Solve for x if
I: Log2(x²-2x+5)=2
ii: Log 3(x²+2x+2)=0
Bravo. Smart you
I assume the base is set to base 2
Log2(x²-2x+5)=2
(x²-2x+5)=2²
x²-2x+5=4
x²-2x+1=0
(x-1)²=0
x=1
Similarly
Am assuming the base is 3
Log 3(x²+2x+2)=0
(x²+2x+2)=3⁰
x²+2x+2=1
x²+2x+1=0
Factored as
(x+1)²=0
x=-1
If they are not to base 10
(x²-2x+5)=10²/2
(x²-2x+5)=50
x²-2x=45
(x-1)²=45+1
(x-1)²=46
not too nice
Same goes for
Log 3(x²+2x+2)=0
x²+2x+2=1/3
x²+2x=1/3-6/3
(x+1)²=-5/3+3/3
(x+1)²=-2/3
Which does work well we have an imaginary result
(x+1)=±i√(2/3)
Log3(x2+2x+2)=0
Hwkajaiakajan
Show me answer
Very good
To solve the equations, we need to isolate the variable x.
For equation I:
Step 1: Start by rewriting the equation to remove the logarithm:
Log2(x²-2x+5) = 2
Step 2: Apply the logarithmic property and rewrite the equation in exponential form:
2 = 2^(Log2(x²-2x+5))
Step 3: Simplify the equation:
2 = x² - 2x + 5
Step 4: Rearrange the equation to a quadratic form:
x² - 2x + 5 - 2 = 0
Step 5: Combine like terms:
x² - 2x + 3 = 0
Step 6: Factorize the quadratic equation:
(x - 1)(x - 3) = 0
Step 7: Set each factor equal to zero and solve for x:
x - 1 = 0 -> x = 1
x - 3 = 0 -> x = 3
So the solutions for Equation I are x = 1 and x = 3.
For equation II:
Step 1: Rewrite the equation to remove the logarithm:
Log3(x²+2x+2) = 0
Step 2: Apply the logarithmic property and rewrite the equation in exponential form:
1 = 3^(Log3(x²+2x+2))
Step 3: Simplify the equation:
1 = x² + 2x + 2
Step 4: Rearrange the equation to a quadratic form:
x² + 2x + 2 - 1 = 0
Step 5: Combine like terms:
x² + 2x + 1 = 0
Step 6: Factorize the quadratic equation:
(x + 1)(x + 1) = 0
Step 7: Set each factor equal to zero and solve for x:
x + 1 = 0 -> x = -1
So the solution for Equation II is x = -1.