what is the solution for the logarithmic equation log2(x2+2x-44)=2?
x2=x squared
8O how can u do that kind of math?! im good but not that good!!
Change your equation to an equivalent exponential equation
so
og2(x2+2x-44)=2 is equivalent to
x^2 + 2x - 44 = 2^2
x^2 + 2x - 48 = 0
(x+8)(x-6)=0
x = -8 or x=6
we have to check both of these answers, since the log function has undefined values.
They both work
To find the solution for the logarithmic equation log2(x^2+2x-44)=2, we need to eliminate the logarithm and solve for x. Here's how you can do it step by step:
Step 1: Rewrite the equation using exponential form. Since the logarithm is to the base 2, we can rewrite the equation as follows:
2^2 = x^2 + 2x - 44
Step 2: Simplify the equation:
4 = x^2 + 2x - 44
Step 3: Rearrange the equation to set it equal to zero:
x^2 + 2x - 48 = 0
Step 4: Factorize the quadratic equation:
(x + 8)(x - 6) = 0
Step 5: Set each factor equal to zero and solve for x:
x + 8 = 0 or x - 6 = 0
Solving the first equation:
x + 8 = 0
x = -8
Solving the second equation:
x - 6 = 0
x = 6
Therefore, the solutions to the logarithmic equation log2(x^2+2x-44)=2 are x = -8 and x = 6.