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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.

  1. \(12 - 4{{\bf{e}}^{7 + 3\,x}} = 7\) Solution
  2. \(1 = 10 - 3{{\bf{e}}^{{z^{\,2}} - 2\,z}}\) Solution
  3. \(2t - t{{\bf{e}}^{6\,t - 1}} = 0\) Solution
  4. \(4x + 1 = \left( {12x + 3} \right){{\bf{e}}^{{x^2} - 2}}\) Solution
  5. \(2{{\bf{e}}^{3\,y + 8}} - 11{{\bf{e}}^{5 - 10\,y}} = 0\) Solution
  6. \(14{{\bf{e}}^{6 - x}} + {{\bf{e}}^{12x - 7}} = 0\) Solution
  7. \(\displaystyle 1 - 8\ln \left( {\frac{{2x - 1}}{7}} \right) = 14\) Solution
  8. \(\ln \left( {y - 1} \right) = 1 + \ln \left( {3y + 2} \right)\) Solution
  9. \(\log \left( w \right) + \log \left( {w - 21} \right) = 2\) Solution
  10. \(2\log \left( z \right) - \log \left( {7z - 1} \right) = 0\) Solution
  11. \(16 = {17^{t - 2}} + 11\) Solution
  12. \({2^{3 - 8w}} - 7 = 11\) 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,

  1. 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.

  2. if interest is compounded continuously we will have, \[A = P{{\bf{e}}^{r\,t}}\]

    dollars after \(t\) years.

  1. 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
    1. quarterly
    2. monthly
    3. continuously
    Solution
  2. 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
    1. quarterly
    2. monthly
    3. continuously
    Solution


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.

  1. A population of bacteria initially has 250 present and in 5 days there will be 1600 bacteria present.
    1. Determine the exponential growth equation for this population.
    2. How long will it take for the population to grow from its initial population of 250 to a population of 2000?
    Solution
  2. We initially have 100 grams of a radioactive element and in 1250 years there will be 80 grams left.
    1. Determine the exponential decay equation for this element.
    2. How long will it take for half of the element to decay?
    3. How long will it take until there is only 1 gram of the element left?
    Solution