Checking logarithm inequality. 
Which one of the following is true.
$(a.)\ \log_{17} 298=\log_{19} 375 \quad  \quad  \quad  \quad (b.)\ \log_{17} 298<\log_{19} 375\\
(c.)\ \log_{17} 298>\log_{19} 375 \quad  \quad  \quad  \quad (d.)\ \text{cannot be determined} $

$17^{2}=289 $ it has a difference of $9$ and $19^{2}=361$ it has a difference of $14$ .
I am not aware of any method if it is there to check such problems,
I would also prefer a method without calculus unless necessary.
I look for a short and simple way .
I have studied maths up to $12$th grade.

Let $x=\log_{17}{298}, y=\log_{19}{375}$.
By definition of logarithms,
$17^x = 298$ and $19^y=375$
So
$17^{x-2} = \dfrac{298}{289} = 1 + \dfrac{9}{289} \tag{1}$ and $19^{y-2}=\dfrac{375}{361} = 1 + \dfrac{14}{361} \tag{2}$.
Now take natural logarithms
$(x-2)\ln{17} = \ln(1+\dfrac{9}{289}) \approx \dfrac{9}{289} \tag{3}$
and
$(y-2)\ln{19} = \ln(1+\dfrac{14}{361}) \approx \dfrac{14}{361} \tag{4}$
From $\ln{19} \approx \ln{17}(1+\frac{2}{17})$ and $\dfrac{14}{361} \times \dfrac{17}{19} \gg \dfrac{9}{289}$
we can say $\dfrac{\frac{14}{361}}{\ln{19}} > \dfrac{\frac{9}{289}}{\ln{17}}$
Then by equations (3), (4)
we have $y-2 > x-2$ or $\boxed{\log_{19}{375} > \log_{17}{298}}$. 
