$\log_9 71$ or $\log_8 61$ I am trying to know  which one is bigger :$$\log_9 71$$ or  $$\log_8  61$$  how can i know without using a calculator ?
 A: $$\log_8 61  = \log_8\left(64\left(1-{3 \over 64}\right)\right) = 2 + \log_8\left(1 - {3 \over 64}\right)$$
$$\log_9 71  = \cdots = 2 + \log_9 \left( 1 - {10 \over 81}\right)$$
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.
A: 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 $$\log_8{61}>\frac{\log_8{71}}{\log_8{9}} = \log_9{71}$$
A: 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:
$\log_b x = \frac{\log_a x} {\log_a b}$
A: 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 $\log_9 71 < \log_8 61$.
A: 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.
A: By drawing tangents of $\log$ curves at points $\log_9 81$ and $\log_8 61$ I obtain the following approximations:
$$\log_9 71 < \log_9 81 - \frac{81-71}{81 \log 9} = 2 - \frac {10}{81 \log 9}$$
$$\log_8 61 > \log_8 64 - \frac{64-61}{61 \log 8} = 2 - \frac {3}{61 \log 8}$$
I can then show that $\frac{10}{81 \log 9} > \frac{3}{61 \log 8}$ as $\frac{10 \cdot 61}{81 \cdot 3} = \frac{610}{243} > 2 > \frac{ \log 9 }{\log 8}$
