Chapter SH, Problem 7.16EP

To determine

To find:the solution for logarithmic equation having base $\text{2}$ .

Answer to Problem 7.16EP

  $\text{x=3}$ .

Explanation of Solution

Given:

  ${\text{4log}}_{\text{2}}{\text{x=log}}_{\text{2}}\text{81}$ .

Concept used:

According the logarithmic formula:

  ${\text{alogx=logx}}^{\text{a}}$ .

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

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

  $\text{x=y}$ .

Calculation:

Here the given equation:

  ${\text{4log}}_{\text{2}}{\text{x=log}}_{\text{2}}\text{81}$ .

According the logarithmic formula:

  ${\text{alogx=logx}}^{\text{a}}$ .

The equation ${\text{4log}}_{\text{2}}{\text{x=log}}_{\text{2}}\text{81}$ can be written as:

  ${\text{log}}_{\text{2}}{\text{x}}^4{\text{=log}}_{\text{2}}\text{81}$ .

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 ${\text{log}}_{\text{2}}{\text{x}}^4{\text{=log}}_{\text{2}}\text{81}$ can be equivalent to:

  ${\text{x}}^4\text{=81}$ .

  $\text{81}$ can be written ${\text{3}}^4$ ,

  ${\text{x}}^4\text{=}{3}^4$ .

  ${\text{x}}^4-3^4=0$ .

  ${\left({\text{x}}^2\right)}^2-{\left(3^2\right)}^2=0$ .

  $\left({\text{x}}^{\text{2}}{\text{+3}}^{\text{2}}\right)\left({\text{x}}^{\text{2}}{\text{-3}}^{\text{2}}\right)\text{=0}$ .

  ${\text{x}}^{\text{2}}{\text{+3}}^{\text{2}}{\text{=0 and x}}^{\text{2}}{\text{-3}}^{\text{2}}\text{=0}$ .

  $\text{x=}\sqrt{\text{-9}}\text{ and x=}\sqrt{\text{9}}$ .

  $\text{x=}\pm \text{3 and x=}\pm \text{3i}$

Since the value of $\text{x }$ is inside the logarithm therefore the value is taken positive and real.

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