Answer
$b=e^{2}$
Work Step by Step
Using the rule $(\log_{a}(g(x)))'=\frac{\frac{g'(x)}{g(x)}}{\ln a}$
So:
$$f'(x)=\frac{\frac{(3x^{2}-2)'}{3x^{2}-2}}{\ln b}$$
$$f'(x)=\frac{\frac{(3x^{2})'-(2)'}{3x^{2}-2}}{\ln b}$$
$$f'(x)=\frac{\frac{3\cdot 2x-0}{3x^{2}-2}}{\ln b}$$
$$f'(x)=\frac{\frac{6x}{3x^{2}-2}}{\ln b}$$
$$f'(1)=\frac{\frac{6\cdot 1}{3\cdot 1^{2}-2}}{\ln b}$$
$$3=\frac{\frac{6\cdot 1}{3\cdot 1^{2}-2}}{\ln b}$$
$$3\ln b=\frac{6\cdot 1}{3\cdot 1^{2}-2}$$
$$3\ln b=\frac{6}{1}$$
$$3\ln b=6$$
$$\ln b=2$$
$$e^{\ln b}=e^{2}$$
$$b=e^{2}$$
