Question

What are the solutions to log3x log3(x2 2) = 1 2log3x? x = –2 x = –1 x = 1 x = 2 There is no true solution.

Answer 1

The solutions of $\log_3x + \log_3(x^2 + 2) = 1 + 2\log_3x$ can be calculated using the logarithm rule.

The solution of $\log_3x + \log_3(x^2 + 2) = 1 + 2\log_3x$  is $x=2$ and $x=1$.

Given:

The expression is,

$\log_3x + \log_3(x^2 + 2) = 1 + 2\log_3x$

What is a logarithm?

The logarithm is the inverse function of exponentiation. In logarithm base must be raised to yield a given number for an exponent. The following rule will use solve the given problem.

• $\log_aa=1$
• $\log_a(mn)=\log_am+\log_bn$
• $\log_ax^n=n\log_ax$

Apply the above rule to solve the given problem.

$\log_3x(x^2+2)=\log_33+\log_3x^2\\ \log_3x(x^2+2)=\log_33x^2\\ x(x^2+2)=3x^2$

Now, solve the above algebraic expression,

$x^2-3x+2=0\\ x^2-2x-x+2=0\\ x(x-2)-1(x-2)=0\\ (x-2)(x-1)=0$

Thus, the solution of $\log_3x + \log_3(x^2 + 2) = 1 + 2\log_3x$  is $x=2$ and $x=1$.

Learn more about the logarithm rule here:

brainly.com/question/7302008

Answer 2

Answer:

x = 1

x = 2

Step-by-step explanation: correct on edg2022