Chapter SH, Problem 7.12EP

To determine

To evaluate: the solution for logarithmic equation to the base $\text{6}$ .

Answer to Problem 7.12EP

  $\text{x=3}$ .

Explanation of Solution

Given:

  ${\text{log}}_{\text{6}}\left(\text{5x-3}\right){\text{=log}}_{\text{6}}\left(\text{x+9}\right)$ .

Concept used:

If the base of the logarithmic is same in both side it can be cancels out:

For example:

  ${\text{log}}_{\text{b}}\left(\text{x}\right){\text{=log}}_{\text{b}}\left(\text{y}\right)$ .

  $\text{x=y}$ .

Calculation:

Here the given equation:

  ${\text{log}}_{\text{6}}\left(\text{5x-3}\right){\text{=log}}_{\text{6}}\left(\text{x+9}\right)$ .

Here the base ofthe logarithmic is same in both side so it can be cancels out:

  ${\text{log}}_{\text{b}}\left(\text{x}\right){\text{=log}}_{\text{b}}\left(\text{y}\right)$ .

  $\text{x=y}$ .

So, the given equation ${\text{log}}_{\text{6}}\left(\text{5x-3}\right){\text{=log}}_{\text{6}}\left(\text{x+9}\right)$ can be equivalent to:

  $\text{5x-3=x+9}$ .

  $\text{4x=12}$ .

Dividing both side by $\text{4}$ :

Hence, $\text{x=3}$ .