TITLE: Checking logarithm inequality.
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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 $\log_{17} 298 < \log_{19} 375.$