If two equivalent logarithmic expressions have the same base, then the arguments must be equal.
x ≈ 13.223
Practice makes perfect
We want to solve an equation involving more than one logarithm.
To do so, we will use the Product Property of Logarithms.
log_b mn = log_b m + log_b n
First, we will isolate the logarithm that contains the variable. Then, we can use the above property to isolate the variable from the logarithm. Note that the property holds also for the natural logarithm.
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^2-2x-e^5=0
⇕
1x^2+( -2)x+( - e^5)=0
We see that a= 1, b= -2, and c= - e^5. 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(1+e^5). Therefore, the exact solutions are x_1=1+sqrt(1+e^5) and x_2=1-sqrt(1+e^5). Let's use a calculator to approximate these values.
x_1 & = 1+sqrt(1+e^5) ≈ 13.223
x_2 & =1-sqrt(1+e^5) ≈ - 11.223
To check for extraneous solutions, we will substitute both 13.223 and - 11.223 for x in the given equation one at a time.
