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If two equivalent logarithmic expressions have the same base, then the arguments must be equal.
x ≈ 10.24
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.
log_6(m) + log_6(n)=log_6(mn)
Distribute (x-3)
m=log_3(3^m)
Calculate power
Next, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal.
We will use the Quadratic Formula to solve the obatained 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^2-12x+18=0 ⇕ 1x^2+( -12)x+ 18=0 We see that a= 1, b= -12, and c= 18. Let's substitute these values into the Quadratic Formula.
Substitute values
- (- a)=a
Calculate power
a * 1=a
Multiply
Subtract term
Split into factors
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root
Factor out 6
a/b=.a /2./.b /2.
Distribute 3
Using the Quadratic Formula, we found that the solutions of the given equation are x=6 ± 3sqrt(2). Therefore, the exact solutions are x_1=6 + 3sqrt(2) and x_2=6 - 3sqrt(2). Let's use a calculator to approximate these values. x_1 & = 6 + 3sqrt(2) ≈ 10.24 x_2 & = 6 - 3sqrt(2) ≈ 1.76 To check for extraneous solutions, we will substitute both 10.24 and 1.76 for x in the given equation one at a time.
x= 10.24
Subtract terms
Calculate logarithm
Add terms
Since substituting 10.24 for x in the given equation produces a true statement, x=10.24 is a solution to our equation. Let's now check for the x=1.76.
x= 1.76
Subtract terms
The argument of a logarithmic function cannot be negative, so both log_3 (-7.24) and log_3 (-1.24) are undefined, and 1.76 is an extraneous solution.