I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 1.9 : Exponential And Logarithm Equations
For problems 1 – 12 find all the solutions to the given equation. If there is no solution to the equation clearly explain why.
- \(12 - 4{{\bf{e}}^{7 + 3\,x}} = 7\) Solution
- \(1 = 10 - 3{{\bf{e}}^{{z^{\,2}} - 2\,z}}\) Solution
- \(2t - t{{\bf{e}}^{6\,t - 1}} = 0\) Solution
- \(4x + 1 = \left( {12x + 3} \right){{\bf{e}}^{{x^2} - 2}}\) Solution
- \(2{{\bf{e}}^{3\,y + 8}} - 11{{\bf{e}}^{5 - 10\,y}} = 0\) Solution
- \(14{{\bf{e}}^{6 - x}} + {{\bf{e}}^{12x - 7}} = 0\) Solution
- \(\displaystyle 1 - 8\ln \left( {\frac{{2x - 1}}{7}} \right) = 14\) Solution
- \(\ln \left( {y - 1} \right) = 1 + \ln \left( {3y + 2} \right)\) Solution
- \(\log \left( w \right) + \log \left( {w - 21} \right) = 2\) Solution
- \(2\log \left( z \right) - \log \left( {7z - 1} \right) = 0\) Solution
- \(16 = {17^{t - 2}} + 11\) Solution
- \({2^{3 - 8w}} - 7 = 11\) Solution
Compound Interest. If we put \(P\) dollars into an account that earns interest at a rate of \(r\) (written as a decimal as opposed to the standard percent) for \(t\) years then,
- if interest is compounded \(m\) times per year we will have,
\[A = P{\left( {1 + \frac{r}{m}} \right)^{t\,m}}\]
dollars after \(t\) years.
- if interest is compounded continuously we will have,
\[A = P{{\bf{e}}^{r\,t}}\]
dollars after \(t\) years.
- We have $10,000 to invest for 44 months. How much money will we have if we put the money into an account that has an annual interest rate of 5.5% and interest is compounded
- quarterly
- monthly
- continuously
- We are starting with $5000 and we’re going to put it into an account that earns an annual interest rate of 12%. How long should we leave the money in the account in order to double our money if interest is compounded
- quarterly
- monthly
- continuously
Exponential Growth/Decay. Many quantities in the world can be modeled (at least for a short time) by the exponential growth/decay equation.
\[Q = {Q_0}{{\bf{e}}^{k\,t}}\]If \(k\) is positive we will get exponential growth and if \(k\) is negative we will get exponential decay.
- A population of bacteria initially has 250 present and in 5 days there will be 1600 bacteria present.
- Determine the exponential growth equation for this population.
- How long will it take for the population to grow from its initial population of 250 to a population of 2000?
- We initially have 100 grams of a radioactive element and in 1250 years there will be 80 grams left.
- Determine the exponential decay equation for this element.
- How long will it take for half of the element to decay?
- How long will it take until there is only 1 gram of the element left?