Solving for unknowns in the exponent
Logarithms are very useful in solving equations where the unknown is in the exponent. First go through the following example and then try the others on your own:
#[Example][Ejemplo]#:
#[Solve for $x$:][Despeja a $x$:]# $\quad 5^{2x} = \dfrac{1}{125}$.
#[How to solve: Take the logarithm of both sides (always works).][Cómo resolver: Toma el logarítmo de ambos lados (siempre funciona).]#
\gap[20] \t $5^{2x} = \dfrac{1}{125}$
\\ \gap[20] \t $\log (5^{2x})=\log \left(\dfrac{1}{125}\right)$
\\ \gap[20] \t $2x\log 5 = -\log 125$
\\ \gap[20] \t $x=-\dfrac{\log 125}{2\log 5}$
In this case, we can again use the algebra of logarithms to get the answer in simpler form:
$x=-\dfrac{\log 125}{2\log 5}$ \t ${}= -\dfrac{\log (5^3)}{2\log 5}$
\\ \t ${}= -\dfrac{3\log 5}{2\log 5}$ \t \gap[40]
\\ \t ${}= -\dfrac{3}{2}$ \t \gap[40]