Chapter SH, Problem 7.15EP

To determine

To find:the solution for logarithmic equation having base $7$ .

Answer to Problem 7.15EP

  $\text{x=3}$ .

Explanation of Solution

Given:

  ${\text{log}}_{\text{7}}{\text{6-log}}_{\text{7}}{\text{2=log}}_{\text{7}}\text{x}$ .

Concept used:

According the logarithmic formula for subtraction of two logarithmic term will be:

  $\text{loga-logb=log}\left(\frac{\text{a}}{\text{b}}\right)$ or $\text{log}\left(\frac{\text{a}}{\text{b}}\right)\text{=loga-logb}$ .

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

  ${\text{log}}_{\text{b}}{\text{x=log}}_{\text{b}}\text{y}$ .

  $\text{x=y}$ .

Calculation:

Here the given equation:

  ${\text{log}}_{\text{7}}{\text{6-log}}_{\text{7}}{\text{2=log}}_{\text{7}}\text{x}$ .

According the logarithmic formula for subtraction of two logarithmic term will be:

  $\text{loga-logb=log}\left(\frac{\text{a}}{\text{b}}\right)$ or $\text{log}\left(\frac{\text{a}}{\text{b}}\right)\text{=loga-logb}$ .

So, from the given equation ${\text{log}}_{\text{7}}{\text{6-log}}_{\text{7}}{\text{2=log}}_{\text{7}}\text{x}$ can be written as:

  ${\text{log}}_{\text{7}}\left(\frac{\text{6}}{2}\right){\text{=log}}_{\text{7}}\text{x}$ .

  ${\text{log}}_{\text{7}}\left(3\right){\text{=log}}_{\text{7}}\text{x}$ .

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

  ${\text{log}}_{\text{b}}{\text{x=log}}_{\text{b}}\text{y}$ .

  $\text{x=y}$ .

So, the given equation ${\text{log}}_{\text{7}}\left(3\right){\text{=log}}_{\text{7}}\text{x}$ can be equivalent to:

  $\text{x=3}$ .

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