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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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Hint : The best way to proceed here is to reduce the equation down to a single exponential.
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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.

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Now 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.