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.
 A: 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 $\boxed{\log_{19}{375} > \log_{17}{298}}$. 
A: The following method does not use approximate calculations.
First of all note that $\ln 19 <\frac{7}{6}\ln 17$.
$$\log_{17} 298 \vee \log_{19} 375$$
$$\frac{\ln 298}{\ln 17}\vee \frac{\ln 375}{\ln 19}$$
$$\frac{\ln 298}{\ln 17}-2\vee \frac{\ln 375}{\ln 19}-2$$
$$\frac{\ln \frac{298}{17^2}}{\ln 17}\vee \frac{\ln \frac{375}{19^2}}{\ln 19}$$
$$\frac{\ln 17}{\ln\frac{298}{17^2}} \overline{\vee} \frac{\ln 19}{\ln\frac{375}{19^2}}$$
Now we use $\ln 19 <\frac{7}{6}\ln 17$ (we will prove that the left number is bigger than the right number)
$$\frac{1}{\ln\frac{298}{17^2}} \overline{\vee} \frac{\frac{7}{6}}{\ln\frac{375}{19^2}}$$
$$\frac{6}{\ln\frac{298}{17^2}} \overline{\vee} \frac{7}{\ln\frac{375}{19^2}}$$
$$6\ln\frac{375}{19^2}\overline{\vee} 7\ln\frac{298}{17^2}$$
$$\left(\frac{375}{19^2} \right )^6\overline{\vee} \left(\frac{298}{17^2} \right )^7$$
It is painfull but possible to calculate without a calculator that $\left(\frac{375}{19^2} \right )^6> \left(\frac{298}{17^2} \right )^7.$ Therefor we have $\log_{17} 298 < \log_{19} 375.$
