Chapter 4.5, Problem 91E

a.

To determine

Solve each equation.

Answer to Problem 91E

  $\boxed{x=1.1}$

Explanation of Solution

Given information:

Each of these equations can be transformed into an equation of linear or quadratic type by applying the hint. Solve each equation.

  ${\left(x-1\right)}^{\log \left(x-1\right)}=100\left(x-1\right)$ [Take log of each side.]

Calculation:

Consider,

  ${\left(x-1\right)}^{\log \left(x-1\right)}=100\left(x-1\right)$

Taking log both sides,

  $\begin{array}{l}{\left(x-1\right)}^{\log \left(x-1\right)}=100\left(x-1\right)\mathrm{.....}\left(1\right)\\ \log {\left(x-1\right)}^{\log \left(x-1\right)}=\log \left\{100\left(x-1\right)\right\}\\ \log \left(x-1\right)\times \log \left(x-1\right)=\log 100+\log \left(x-1\right)\\ {\left\{\log \left(x-1\right)\right\}}^2=\log {\left(10\right)}^2+\log \left(x-1\right)\\ {\left\{\log \left(x-1\right)\right\}}^2=2{\log }_{10}\left(10\right)+\log \left(x-1\right)\\ {\left\{\log \left(x-1\right)\right\}}^2=2\times 1+\log \left(x-1\right)\\ {\left\{\log \left(x-1\right)\right\}}^2=2+\log \left(x-1\right)\\ {\left\{\log \left(x-1\right)\right\}}^2-\log \left(x-1\right)-2=\mathrm{0.....}\left(2\right)\end{array}$

Put $\log \left(x-1\right)=a$

  $\begin{array}{l}{a}^2-a-2=0\\ factorize\\ a^2-2a+a-2=0\\ a\left(a-2\right)+1\left(a-2\right)\\ \left(a-2\right)\left(a+1\right)=0\\ \left(a-2\right)=0,a=2\\ \left(a+1\right)=0,a=-1\end{array}$

  $\begin{array}{l}{\log }_{10}\left(x-1\right)=2\\ \left(x-1\right)={10}^2\\ \left(x-1\right)=100\\ x=101\\ and\\ a=-1,then\\ {\log }_{10}\left(x-1\right)=-1\\ x-1={10}^{-1}\\ x-1=\frac{1}{10}\\ x=1+\frac{1}{10},x=\frac{11}{10}\end{array}$

  $\boxed{x=1.1}$

Hence, the solution is, $\boxed{x=1.1}$

b.

To determine

Solve each equation.

Answer to Problem 91E

  $\boxed{x=64}$

Explanation of Solution

Given information:

Each of these equations can be transformed into an equation of linear or quadratic type by applying the hint. Solve each equation.

  ${\log }_{2}x+{\log }_{4}x+{\log }_{8}x=11$ [Change all logs to base $2$ .]

Calculation:

Consider,

  $\begin{array}{l}lo{g}_{2}x+lo{g}_{4}x+lo{g}_{8}x=11\\ \frac{logx}{log2}+\frac{logx}{log4}+\frac{logx}{log8}=11\left[\therefore {\log }_{a}b=\frac{\log b}{\log a}\right]\\ \frac{logx}{log2}+\frac{logx}{log{2}^2}+\frac{logx}{log{2}^3}=11\\ \frac{logx}{log2}+\frac{logx}{log{2}^2}+\frac{logx}{log{2}^3}=11\\ \frac{logx}{log2}+\frac{logx}{2log2}+\frac{logx}{3log2}=11\\ \frac{logx}{log2}+\frac{logx}{2log2}+\frac{logx}{3log2}=11\\ lo{g}_{2}x+\frac{\mathrm{l}}{2}lo{g}_{2}x+\frac{1}{3}lo{g}_{2}x=11\\ lo{g}_{2}x\left(1+\frac{\mathrm{l}}{2}+\frac{1}{3}\right)=11\\ lo{g}_{2}x\left(\frac{1\mathrm{l}}{6}\right)=11\\ lo{g}_{2}x=6\\ x=2^6\end{array}$

  $\boxed{x=64}$

Hence, the solution is, $\boxed{x=64}$

c.

To determine

Solve each equation.

Answer to Problem 91E

  $\boxed{x={\log }_{2}3}$

Explanation of Solution

Given information:

Each of these equations can be transformed into an equation of linear or quadratic type by applying the hint. Solve each equation.

  $4^3-2^{x+1}=3$ [Write as a quadratic in $2^x$ ]

Calculation:

Consider,

  $4^3-2^{x+1}=3$

  $\begin{array}{l}{\left(2\right)}^{2x}-{\left(2\right)}^x.{\left(2\right)}^1=3\left[\therefore a^{b+c}=a^b.a^c\right]\\ {\left(2\right)}^{2x}-2{\left(2\right)}^x-3=0\end{array}$

Now, this is a quadratic equation, put $2^x=a$

  $\begin{array}{l}{a}^2-2a-3=0\\ a^2-3a+a-3=0\\ a\left(a-3\right)+1\left(a-3\right)=0\\ \left(a-3\right)\left(a+1\right)=0\\ \left(a-3\right)=0,\left(a+1\right)=0\\ a=3,a=-1\end{array}$

So, $2^x=a$

  $\begin{array}{l}{2}^x=3\\ {\log }_{2}3=x\\ 2^x=-1\left(notdefined\right)\end{array}$

  $\boxed{x={\log }_{2}3}$

Hence, the solution is, $\boxed{x={\log }_{2}3}$