# 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.

• I wonder if a typo happened here. Might someone have intended $\log_{17}289$? Since $17^2=289$, we get $\log_{17}289=2$. Then we could say that since $19^2=361$ we have $2 = \log_{19}361 < \log_{19} 375$. ${}\qquad{}$ – Michael Hardy Jul 26 '15 at 23:07
• No typo there . – R K Jul 26 '15 at 23:09
• $\log_{18} 335$ and $\log_{18} 336$ are between the two numbers. Does this help? – peterwhy Jul 26 '15 at 23:13
• This can of course be done with a calculator, but I can't help suspecting that some intelligent method was intended to be used instead. – Michael Hardy Jul 26 '15 at 23:22
• @peterwhy : Did you conclude that $\log_{18}335$ is between these two simply by numerical computation (in which case, why bother with so indirect an approach?) or do you have some intelligent reason to draw that conclusion? ${}\qquad{}$ – Michael Hardy Jul 26 '15 at 23:31

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 $\boxed{\log_{19}{375} > \log_{17}{298}}$.

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 $\log_{17} 298 < \log_{19} 375.$

• You might want to explain why $\ln 19 <\frac{7}{6}\ln 17$ without using the $\ln$ function directly. – Marconius Aug 7 '15 at 13:20
• It is not a problem. $$\ln 19 \vee \frac{7}{6}\ln 17$$ $$6\ln 19 \vee 7\ln 17$$ $$\ln 19^6 \vee \ln 17^7$$ $$19^6<17^7$$ – Tzara_T'hong Aug 7 '15 at 13:29