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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
5. Find all the solutions to \(2{{\bf{e}}^{3\,y + 8}} - 11{{\bf{e}}^{5 - 10\,y}} = 0\). If there are no solutions clearly explain why.
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With both exponentials in the equation this may be a little difficult to solve, so let’s do some work to reduce this down to an equation with a single exponential.
\[\begin{align*}2{{\bf{e}}^{3\,y + 8}} & = 11{{\bf{e}}^{5 - 10\,y}}\\ \frac{{{{\bf{e}}^{3\,y + 8}}}}{{{{\bf{e}}^{5 - 10\,y}}}} & = \frac{{11}}{2}\\ {{\bf{e}}^{13\,y + 3}} & = \frac{{11}}{2}\end{align*}\]Note that we could have divided by either exponential but by dividing by the one that we did we avoid a negative exponent on the \(y\), which is sometimes easy to lose track of.
Show Step 2Now all we need to do is take the logarithm of both sides and solve for \(y\).
\[\begin{align*}\ln \left( {{{\bf{e}}^{13\,y + 3}}} \right) & = \ln \left( {\frac{{11}}{2}} \right)\\ 13y + 3 & = \ln \left( {\frac{{11}}{2}} \right)\\ y & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{13}}\left( {\ln \left( {\frac{{11}}{2}} \right) - 3} \right) = - 0.09963}}\end{align*}\]Depending upon your preferences either the exact or decimal solution can be used.