Respuesta :
Answer:
Option C and D are correct
x = 1
x= 2
Step-by-step explanation:
Use logarithmic rules:
[tex]\log_b b = 1[/tex]
[tex]\log_b (mn) = \log_b m+ \log_b n[/tex]
[tex]\log_b x^n = n\log_b x[/tex]
As per the statement:
Given the equation:
[tex]\log_3 x+ \log_3 (x^2+2) = 1+2\log_3 x[/tex]
Apply the logarithmic rules:'
[tex]\log_3 x(x^2+2) = \log_3 3 + \log_3 x^2[/tex]
Again apply logarithmic rule:
[tex]\log_3 x(x^2+2) =\log_3 3x^2[/tex]
then;
[tex]x(x^2+2) = 3x^2[/tex]
Divide both sides by x we have;
[tex]x^2+2 = 3x[/tex]
Subtract 3x from both sides we have;
[tex]x^2-3x+2=0[/tex]
⇒[tex]x^2-2x-x+2=0[/tex]
⇒[tex]x(x-2)-1(x-2)=0[/tex]
⇒[tex](x-2)(x-1)=0[/tex]
By zero product property we have;
x-2 = 0 and x-1 = 0
⇒x = 2 and x = 1
Therefore, the solution for the given equation are: 1 and 2
The solutions of the equation log3x + log3(x2 + 2) = 1 + 2log3x are x =2 and x=1.
Given:
The expression is,
log3x + log3(x2 + 2) = 1 + 2log3x
What is the logarithmic function?
The logarithmic function is the inverse of the exponential function.
In logarithm base must be raised to yield a given number for an exponent.
The following rule will use solve the given problem.
[tex]\rm log_aa=1\\\\loa_amn=log_am+log_an\\\\log_ax^n=n log_ax[/tex]
Therefore,
The solutions of the equations are;
[tex]\rm log_3x + log3(x^2 + 2) = 1 + 2log3x\\\\ \text{ Taking log base 3 on both the sides}\\\\log_3(x \times (x^2+2))=log_33x^2\\\\ x(x^2+2)=3x^2\\\\x^3+2x=3x^2\\\\x^2+2=3x\\\\x^2-3x+2=0\\\\x^2-2x-x+2=0\\\\x(x-2)-1(x-2)=0\\\\(x-2)(x-1)=0\\\\x-2=0, \ x=2\\\\x-1=0, \ x=1[/tex]
Hence, the solutions of the equation log3x + log3(x2 + 2) = 1 + 2log3x are x =2 and x=1.
To learn more about the logarithm rule here:
brainly.com/question/7302008
