Rewrite the logarithmic equation in exponential form.
x = 1+sqrt(41)2 ≈ 3.702 and x = 1-sqrt(41)2 ≈ -2.702
Practice makes perfect
To solve the given logarithmic equation, we will rewrite it in exponential form using the definition of a logarithm.
log_b x=y ⇔ x= b^y
This definition tells us how to rewrite the logarithm equivalent of y in exponential form. The argument x is equal to b raised to the power of y.
log_2( x^2-x-6)=2
⇕
x^2-x-6= 2^2We can see above that 2 is the exponent to which 2 must be raised to obtain x^2-x-6. Now, let's simplify our equation.
We will use the Quadratic Formula to solve the given quadratic equation.
ax^2+ bx+ c=0 ⇔ x=- b± sqrt(b^2-4 a c)/2 a
We first need to identify the values of a, b, and c.
x^2-x-10=0
⇕
1x^2+( - 1)x+( - 10)=0
We see that a= 1, b= - 1, and c= - 10. Let's substitute these values into the Quadratic Formula.
Using the Quadratic Formula, we found that the solutions of the given equation are x= 1± sqrt(41)2. Therefore, the exact solutions are x_1= 1+sqrt(41)2 and x_2= 1-sqrt(41)2. Let's use a calculator to approximate these values.
x_1 & = 1+sqrt(41)2 ≈ 3.702
x_2 & = 1-sqrt(41)2 ≈ - 2.702
To check our answer, we will substitute both 3.702 and - 2.702 for x in the given equation one at a time.
Since substituting 3.702 for x in the given equation produces a true statement, x≈ 3.702 is a solution to our equation. Let's now check for the x≈ -2.702.
