## College Algebra (11th Edition)

$x=-2$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $x=\log_{4/5} \dfrac{25}{16} ,$ use the properties of logarithms. $\bf{\text{Solution Details:}}$ Using exponents, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\log_{4/5} \dfrac{5^2}{4^2} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\log_{4/5} \dfrac{4^{-2}}{5^{-2}} .\end{array} Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\log_{4/5} \left(\dfrac{4}{5}\right)^{-2} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} x=-2\log_{4/5} \dfrac{4}{5} .\end{array} Since $\log_b b =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=-2(1) \\\\ x=-2 .\end{array}