## Precalculus (6th Edition) Blitzer

$x \approx 1.69$
The base in the exponential equation is $19$, so take the natural logarithm on both sides to obtain $\ln{19^x}=\ln{143}.$ Use the power rule $\ln{b^x} = x \ln{b}$ to bring the exponent to the front: $x \ln{19} = \ln{143}.$ Divide both sides of the equation by $\ln{19}$ to obtain $x=\dfrac{\ln{143}}{\ln{19}}.$ Use a calculator and round-off the answer to two decimal places to obtain $x \approx 1.69.$