1) We must solve for x the following equation:
[tex]e^{5x}=25.[/tex]To solve this equation, we take the natural logarithm to both sides of the equation:
[tex]\begin{gathered} \ln (e^{5x})=\ln (25), \\ 5x\cdot\ln e=\ln 25. \end{gathered}[/tex]Now, we use the following results:
[tex]\begin{gathered} \ln e=1, \\ \ln (25)=\ln (5^2)=2\cdot\ln 5. \end{gathered}[/tex]Replacing these results in the equation above, we have:
[tex]5x=2\cdot\ln 5.[/tex]Solving for x, we get:
[tex]x=\frac{2}{5}\cdot\ln 5\cong0.64.[/tex]2) We must solve for x the following equation:
[tex]\ln (2x)+\ln (7)=4.[/tex]To solve this problem, we isolate the part that involves the x:
[tex]\ln (2x)=4-\ln (7)\text{.}[/tex]Now, using the following property:
[tex]\ln y=z\rightarrow y=e^z\text{.}[/tex]with:
[tex]\begin{gathered} y=2x, \\ z=4-\ln 7. \end{gathered}[/tex]we have:
[tex]\ln (2x)=4-\ln 7\rightarrow2x=e^{4-\ln 7}.[/tex]Solving the last equation for x, we get:
[tex]x=\frac{1}{2}\cdot e^{4-\ln 7}\cong3.90.[/tex]Answers
1) The value of x that solves the first equation is 0.64 to two decimal places.
2) The value of x that solves the second equation is 3.90 to two decimal places.
Review of the base of a logarithm
We can define the logarithm in base a through the following equations:
[tex]\begin{gathered} \log _aa=1, \\ \log _aa^x=x\cdot\log _aa=x\cdot1=x\text{.} \end{gathered}[/tex]When we use as a base the Euler number e ≅ 2.718, the logarithm is called "natural" and we use the following notation for it:
[tex]_{}\log _e=\ln .[/tex]With this notation, we have the following properties:
[tex]\begin{gathered} \ln e=1, \\ \ln e^x=x\cdot\ln e=x\cdot1=x\text{.} \end{gathered}[/tex]