Respuesta :

Posted to your other question as well.

So we have:

A) [tex]\log_x(36)=2[/tex]

B) [tex]\log_3(2x-9)=3[/tex]

C) [tex]\log_3(216)=x[/tex]

D) [tex]\log_3(-2x-3)=2[/tex]

We need to try x = –6 in each equation until we find a solution that makes both sides of the equation equal.

For A, this gives an undefined result. Logarithms are only defined for bases that are positive real numbers not equal to 1.

For B, this again gives an undefined result. The logarithm of a number is only defined when that number is a positive real number. It would be –21, which is undefined.

For C, we have a defined operation, but [tex]\log_3(216)=4.893 \neq -6[/tex].

Finally, for D, we have 

[tex]\log_3(-2(-6)-3)=2\\ \log_3(9)=2 \\ 3^2=9 \\ 9=9[/tex]

This is true.

The equation that has x =-6 as a solution is [tex]log_3(-2x-3)=2[/tex]

  • The indices equivalent of the logarithmic function [tex]log_ab=x[/tex] is expressed as [tex]a^x=b[/tex]

  • For the logarithmic equation [tex]log_3(-2x-3)=2[/tex], this can be expressed as [tex]3^2=-2x-3[/tex]

Simplify to check whether x = 6;

[tex]9= -2x - 3\\-2x = 9 + 3\\-2x = 12\\x = -12/2\\x = -6[/tex]

Hence the equation that has x =-6 as a solution is [tex]log_3(-2x-3)=2[/tex]

Learn more on logarithm here: https://brainly.com/question/12049968