## College Algebra (11th Edition)

$x\approx -5.057$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $3^{x-4}=7^{2x+5} ,$ take the logarithm of both sides. Then use the properties of logarithms and the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log3^{x-4}=\log7^{2x+5} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x-4)\log3=(2x+5)\log7 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(\log3)-4(\log3)=2x(\log7)+5(\log7) \\\\ x\log3-4\log3=2x\log7+5\log7 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x\log3-2x\log7=5\log7+4\log3 \\\\ x(\log3-2\log7)=5\log7+4\log3 \\\\ x=\dfrac{5\log7+4\log3}{\log3-2\log7} \\\\ x\approx -5.057 .\end{array}