Solution
$$y = a + b \log_e x$$
=> $$b\log_ex=y-a$$
=> $$\log_ex^b=y-a$$
=> $$x^b=e^{\left(y-a\right)}$$
=> $$x^b=\dfrac{e^y}{e^a}$$
=> $$e^y=e^ax^b$$
Thus, we can clearly see that $$e^y$$ is directly proportional to $$x^b$$
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