## College Algebra (11th Edition)

$x\approx 3.240$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $6^{x+1}=4^{2x-1} ,$ 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} \log6^{x+1}=\log4^{2x-1} .\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+1)\log6=(2x-1)\log4 .\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(\log6)+1(\log6)=2x(\log4)-1(\log4) \\\\ x\log6+\log6=2x\log4-\log4 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x\log6-2x\log4=-\log4-\log6 \\\\ x(\log6-2\log4)=-\log4-\log6 \\\\ x=\dfrac{-\log4-\log6}{\log6-2\log4} \\\\ x\approx 3.240 .\end{array}