Problem 2)
[tex]200e^{0.3x}=300[/tex]
Divide both sides by 200
[tex]\frac{200e^{0.3x}}{200}= \frac{300}{200}[/tex] ⇒ [tex]e^{0.3x}= \frac{3}{2}[/tex]
[tex]ln(e^{0.3x})=ln(\frac{3}{2})[/tex]
then the following log rule is used on the left side: [tex]\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)[/tex]
[tex]0.3x\ln \left(e\right)=\ln \left(\frac{3}{2}\right)[/tex]
since [tex]ln(e)=1[/tex] we get:
[tex]0.3x=ln(\frac{3}{2})[/tex]
To isolate x, we then multiply both sides by 10 and afterwards divide both sides by 3.
[tex]0.3x*10=ln(\frac{2}{3})*10[/tex] ⇒ [tex]3x=ln(\frac{2}{3})*10[/tex] ⇒ [tex]\frac{3x}{3}= \frac{ln(\frac{2}{3})*10}{3}[/tex] ⇒ [tex]x= \frac{ln(\frac{2}{3})*10}{3} {\approx} -1.35155[/tex]
Problem 3)
[tex]6e^{2x}=7e^{4x}[/tex]
⇒ [tex]ln(6e^{2x})=ln(7e^{4x})[/tex]
then the following log rule is used on both sides: [tex]\log _c\left(ab\right)=\log _c\left(a\right)+\log _c\left(b\right)[/tex]
[tex]\ln \left(6\right)+\ln \left(e^{2x}\right)=\ln \left(7\right)+\ln \left(e^{4x}\right)[/tex]
the following log rule is used on both sides: [tex]\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)[/tex]
[tex]\ln \left(6\right)+2x\ln \left(e\right)=\ln \left(7\right)+4x\ln \left(e\right)[/tex]
since [tex]ln(e)=1[/tex] we get:
[tex]\ln \left(6\right)+2x=\ln \left(7\right)+4x[/tex]
To isolate x we subtract [tex]ln(6)[/tex] from both sides:
[tex]\ln \left(6\right)+2x-\ln \left(6\right)=\ln \left(7\right)+4x-\ln \left(6\right)[/tex]
then we can use the following log rule on the right side: [tex]\log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)[/tex]
[tex]2x=\ln \left(\frac{7}{6}\right)+4x[/tex]
then subtract 4x from both sides:
[tex]2x-4x=\ln \left(\frac{7}{6}\right)+4x-4x[/tex] ⇒ [tex]-2x=\ln \left(\frac{7}{6}\right)[/tex]
Divide both sides by -2
[tex]\frac{-2x}{-2}=\frac{\ln \left(\frac{7}{6}\right)}{-2}[/tex] ⇒ [tex]x=-\frac{\ln \left(\frac{7}{6}\right)}{2} {\approx} -0.0771[/tex]
