## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x\approx2.810$
Using the properties of equality, the given equation, $4.9^x-87=0 ,$ is equivalent to \begin{array}{l}\require{cancel} 4.9^x=87 .\end{array} Taking the natural logarithm of both sides and using the properties of logarithms, the value of the variable that satisfies the equation, $4.9^x=87 ,$ is \begin{array}{l}\require{cancel} \ln4.9^x=\ln87 \\\\ x\ln4.9=\ln87 \\\\ x=\dfrac{\ln87}{\ln4.9} \\\\ x\approx2.810 .\end{array}