Chapter 8.6, Problem 4BCYP

To determine

To find: the solution of the inequality and round off to nearest ten-thousand.

Answer to Problem 4BCYP

The value of $y$ must be greater than equal to $-0.875$ , $\mathrm{Im}\left(y\right)=\frac{\pi \left(2n+1\right)}{2\log \left(2\right)}$ and $\mathrm{Im}\left(y\right)=\frac{\pi n}{\log \left(5\right)}.$

Explanation of Solution

Given:

  $4^y<5^{2y+1}.$

Concept used:

In exponential form using the definition of the logarithm. If $\text{x and b }$ are positive real numbers and $\text{b}$ does not equal to $\text{1}$ ,then:

  ${\text{log}}_{\text{b}}\left(\text{x}\right)\text{=y}$ which is equivalent to the ${\text{b}}^{\text{y}}\text{=x}$ .

  $\log x^a=a\log x.$

The value or domain inside the logarithm is always greater than $0$ .

Calculation:

According to the given:

  $4^y<5^{2y+1}.$

Multiplying the both side by logarithmic function:

  $log{4}^y<\log 5^{2y+1}.$

According to the law of logarithm:

  $\log x^a=a\log x.$

There for the inequality equation will be:

  $\begin{array}{l}ylog\left(4\right)<\left(2y+1\right)\log \left(5\right).\\ 0.6y<1.4y+\mathrm{0.7.}\\ 0.6y-1.4y<\mathrm{0.7.}\\ -0.8y<\mathrm{0.7.}\\ y>-\mathrm{0.875.}\end{array}$

The value inside the logarithm should be greater than $0$ .

  $\begin{array}{l}{4}^y>0.\\ \mathrm{Im}\left(y\right)=\frac{\pi \left(2n+1\right)}{2\log \left(2\right)}.\end{array}$

Similarly, the R.H.S can be taken as:

  $\begin{array}{l}{5}^{2y+1}>0.\\ {25}^{y+1}>0.\\ \mathrm{Im}\left(y\right)=\frac{\pi n}{\log \left(5\right)}.\end{array}$

Hence, $y$ must be greater than equal to $-0.875$ , $\mathrm{Im}\left(y\right)=\frac{\pi \left(2n+1\right)}{2\log \left(2\right)}$ and $\mathrm{Im}\left(y\right)=\frac{\pi n}{\log \left(5\right)}.$