[latex]\begin{array}{l} {2}^{x - 1}={2}^{2x - 4}\hfill & \text{The common base is }2.\hfill \\ \text{ }x - 1=2x - 4\hfill & \text{By the one-to-one property the exponents must be equal}.\hfill \\ \text{ }x=3\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

[latex]\begin{array}{l}\text{ }{8}^{x+2}={16}^{x+1}\hfill & \hfill \\ {\left({2}^{3}\right)}^{x+2}={\left({2}^{4}\right)}^{x+1}\hfill & \text{Write }8\text{ and }16\text{ as powers of }2.\hfill \\ \text{ }{2}^{3x+6}={2}^{4x+4}\hfill & \text{To take a power of a power, multiply exponents}.\hfill \\ \text{ }3x+6=4x+4\hfill & \text{Use the one-to-one property to set the exponents equal}.\hfill \\ \text{ }x=2\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

[latex]\begin{array}{l}{2}^{5x}={2}^{\frac{1}{2}}\hfill & \text{Write the square root of 2 as a power of }2.\hfill \\ 5x=\frac{1}{2}\hfill & \text{Use the one-to-one property}.\hfill \\ x=\frac{1}{10}\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

This equation has no solution. There is no real value of x that will make the equation a true statement because any power of a positive number is positive.

Analysis of the Solution

The figure below shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

[latex]\begin{array}{l}\text{ }{5}^{x+2}={4}^{x}\hfill & \text{There is no easy way to get the powers to have the same base}.\hfill \\ \text{ }\mathrm{ln}{5}^{x+2}=\mathrm{ln}{4}^{x}\hfill & \text{Take ln of both sides}.\hfill \\ \text{ }\left(x+2\right)\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use laws of logs}.\hfill \\ \text{ }x\mathrm{ln}5+2\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the distributive law}.\hfill \\ \text{ }x\mathrm{ln}5-x\mathrm{ln}4=-2\mathrm{ln}5\hfill & \text{Get terms containing }x\text{ on one side, terms without }x\text{ on the other}.\hfill \\ x\left(\mathrm{ln}5-\mathrm{ln}4\right)=-2\mathrm{ln}5\hfill & \text{On the left hand side, factor out an }x.\hfill \\ \text{ }x\mathrm{ln}\left(\frac{5}{4}\right)=\mathrm{ln}\left(\frac{1}{25}\right)\hfill & \text{Use the laws of logs}.\hfill \\ \text{ }x=\frac{\mathrm{ln}\left(\frac{1}{25}\right)}{\mathrm{ln}\left(\frac{5}{4}\right)}\hfill & \text{Divide by the coefficient of }x.\hfill \end{array}[/latex]

[latex]\begin{array}{l}100\hfill & =20{e}^{2t}\hfill & \hfill \\ 5\hfill & ={e}^{2t}\hfill & \text{Divide by the coefficient of the power}\text{.}\hfill \\ \mathrm{ln}5\hfill & =2t\hfill & \text{Take ln of both sides}\text{. Use the fact that }\mathrm{ln}\left(x\right)\text{ and }{e}^{x}\text{ are inverse functions}\text{.}\hfill \\ t\hfill & =\frac{\mathrm{ln}5}{2}\hfill & \text{Divide by the coefficient of }t\text{.}\hfill \end{array}[/latex]

Analysis of the Solution

Using laws of logs, we can also write this answer in the form [latex]t=\mathrm{ln}\sqrt{5}[/latex]. If we want a decimal approximation of the answer, we use a calculator.

[latex]\begin{array}{l}4{e}^{2x}+5=12\hfill & \hfill \\ 4{e}^{2x}=7\hfill & \text{Combine like terms}.\hfill \\ {e}^{2x}=\frac{7}{4}\hfill & \text{Divide by the coefficient of the power}.\hfill \\ 2x=\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Take ln of both sides}.\hfill \\ x=\frac{1}{2}\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Solve for }x.\hfill \end{array}[/latex]

[latex]\begin{array}{l}{e}^{2x}-{e}^{x}\hfill & =56\hfill & \hfill \\ {e}^{2x}-{e}^{x}-56\hfill & =0\hfill & \text{Get one side of the equation equal to zero}.\hfill \\ \left({e}^{x}+7\right)\left({e}^{x}-8\right)\hfill & =0\hfill & \text{Factor by the FOIL method}.\hfill \\ {e}^{x}+7\hfill & =0\text{ or }{e}^{x}-8=0 & \text{If a product is zero, then one factor must be zero}.\hfill \\ {e}^{x}\hfill & =-7{\text{ or e}}^{x}=8\hfill & \text{Isolate the exponentials}.\hfill \\ {e}^{x}\hfill & =8\hfill & \text{Reject the equation in which the power equals a negative number}.\hfill \\ x\hfill & =\mathrm{ln}8\hfill & \text{Solve the equation in which the power equals a positive number}.\hfill \end{array}[/latex]

Analysis of the Solution

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation [latex]{e}^{x}=-7[/latex] because a positive number never equals a negative number. The solution [latex]x=\mathrm{ln}\left(-7\right)[/latex] is not a real number, and in the real number system this solution is rejected as an extraneous solution.