## College Algebra (11th Edition)

$x\approx0.823$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $2^{x+3}=5^{2x} ,$ 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} \log2^{x+3}=\log5^{2x} .\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+3)\log2=2x\log5 .\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(\log2)+3(\log2)=2x\log5 \\\\ x\log2+3\log2=2x\log5 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x\log2-2x\log5=-3\log2 \\\\ x(\log2-2\log5)=-3\log2 \\\\ x=-\dfrac{3\log2}{\log2-2\log5} \\\\ x\approx0.823 .\end{array}