$\log_9 71$ or $\log_8 61$

Solution 1:

Ok, so we have that $$\frac{61}{64}>\frac{71}{81} \implies \log_{8}\left(\frac{61}{64}\right)>\log_{9}\left(\frac{71}{81}\right)\cdot\log_{8}(9)>\log_{9}\left(\frac{71}{81}\right)$$ By application of the change of base formula and the fact that $\log_{8}(9)>1$, which is trivial.

Solution 2:

$$\log_8 61 = \log_8\left(64\left(1-{3 \over 64}\right)\right) = 2 + \log_8\left(1 - {3 \over 64}\right)$$

$$\log_9 71 = \cdots = 2 + \log_9 \left( 1 - {10 \over 81}\right)$$

Let's drop the 2s, and note that both values are negative.

$$\log_8\left(1 - {3 \over 64}\right) = {\log(1 - 3/64) \over \log 8} = { 2 \log (1 - 3/64) \over 2 \log 8 } = {\log\left(\left(1-{3 \over 64}\right)^2\right) \over \log 64}$$

Now note that

$$\left(1-{3 \over 64}\right)^2 > 1 - {6 \over 64} > 1 - {6.4 \over 64} = 1 - {8.1 \over 81} > 1 - {10 \over 81}$$

so

$${\log\left(\left(1-{3 \over 64}\right)^2\right) \over \log 64} > {\log \left( 1 - {10 \over 81} \right) \over \log 64} > {\log \left( 1 - {10 \over 81} \right) \over \log 9} = \log_9 \left( 1 - {10 \over 81} \right)$$

Thus, $\log_8(61)$ is greater.

Solution 3:

Notice that $\log_8:(0,\infty) \to \mathbb{R}$ is strictly concave. This means that we have $$\log_8{\left( \frac{x+y}{2}\right)} > \frac{\log_8{x}+\log_8{y}}{2} \implies \log_8{\left( \frac{x+y}{2}\right)^2} > \log_8{(xy)}$$

Since $\log_8{9} >1$, we can conclude that $$\log_8{\left( \frac{x+y}{2}\right)^2} > \frac{\log_8{(xy)}}{ \log_8{9}}$$

Let $x = \sqrt{61}-i\sqrt{10}$ and $y = \sqrt{61}+i\sqrt{10}$. Plugging these in (this is okay because imaginary parts disappear), we find that $$\log_8{61}>\frac{\log_8{71}}{\log_8{9}} = \log_9{71}$$

Solution 4:

If we are allowed to use calculus, we can get a somewhat formal estimate of both numbers: $$\log_9 (71) = \log_9 (81) + \log_9(71/81) = 2 + \log_9 \left(1 - \frac{10}{81}\right) \approx 2 - \frac{1}{2 \ln 3}\left(\frac{10}{81}\right), \\ \log_8 (61) = \log_8 (64) + \log_8 (61/64) = 2 + \log_8 \left(1 - \frac{3}{64}\right) \approx 2 - \frac{1}{3 \ln 2}\left(\frac{3}{64}\right).$$ Using rough estimates like $\ln 2 \approx 0.7$, $\ln 3 \approx 1.1$, $\frac{10}{81} \approx \frac{1}{8} = 0.125$ and $\frac{3}{64} \approx \frac{3}{60} = 0.05$ we get $$\log_9 (71) \approx 2 - \frac{1}{2.1}(0.125) \approx 2 - 0.06 \approx 1.94, \qquad (\text{exact: } \log_9 (71) = 1.940\ldots) \\ \log_8(61) \approx 2 - \frac{1}{2.2}(0.05) \approx 2 - 0.025 \approx 1.975. \qquad (\text{exact: } \log_9(61) = 1.976\ldots)$$ So $\log_9 71 < \log_8 61$.

Solution 5:

You can change the base of the logarithm and put both of them in the same base and then you know that log whose base is bigger than 1 are crescent so u can easily find what's the biggest one.
Change formula: $\log_b x = \frac{\log_a x} {\log_a b}$