The problem asks us to identify which of the four given inequalities is correct. The inequalities involve logarithmic and exponential expressions. The options are: [A] $log_4 7 > log_4 8$ [B] $log_{0.2} 7 > log_{0.2} 8$ [C] $2.1^{\frac{4}{5}} > 3.4^{\frac{4}{5}}$ [D] $0.7^{-2} > 0.7^{-3}$

1. Problem Description

The problem asks us to identify which of the four given inequalities is correct. The inequalities involve logarithmic and exponential expressions. The options are:

[A] $log_4 7 > log_4 8$

[B] $log_{0.2} 7 > log_{0.2} 8$

[C] $2.1^{\frac{4}{5}} > 3.4^{\frac{4}{5}}$

[D] $0.7^{-2} > 0.7^{-3}$

2. Solution Steps

We analyze each option:

[A] $log_4 7 > log_4 8$

Since the base 4 is greater than 1, the logarithmic function is increasing. Therefore, if $7 < 8$, then $log_4 7 < log_4 8$. Thus, $log_4 7 > log_4 8$ is false.

[B] $log_{0.2} 7 > log_{0.2} 8$

Since the base 0.2 is between 0 and 1, the logarithmic function is decreasing. Therefore, if $7 < 8$, then $log_{0.2} 7 > log_{0.2} 8$. Thus, the inequality is true.

[C] $2.1^{\frac{4}{5}} > 3.4^{\frac{4}{5}}$

Since the exponent $\frac{4}{5}$ is positive, the function $x^{\frac{4}{5}}$ is increasing for $x > 0$. Since $2.1 < 3.4$, then $2.1^{\frac{4}{5}} < 3.4^{\frac{4}{5}}$. Thus, $2.1^{\frac{4}{5}} > 3.4^{\frac{4}{5}}$ is false.

[D] $0.7^{-2} > 0.7^{-3}$

We can rewrite these as:

$0.7^{-2} = (\frac{7}{10})^{-2} = (\frac{10}{7})^2 = \frac{100}{49}$

$0.7^{-3} = (\frac{7}{10})^{-3} = (\frac{10}{7})^3 = \frac{1000}{343}$

Comparing the two:

$\frac{100}{49} ? \frac{1000}{343}$

$\frac{100}{49} ? \frac{1000}{7*49}$

$\frac{100}{49} ? \frac{1000/7}{49}$

Comparing $100$ and $1000/7$, we have $100 = 700/7 < 1000/7$.

Therefore, $\frac{100}{49} < \frac{1000}{343}$. Thus, $0.7^{-2} > 0.7^{-3}$ is false.

Alternatively, since $0 < 0.7 < 1$, the function $f(x) = 0.7^x$ is a decreasing function. So, if $-2 > -3$, then $0.7^{-2} < 0.7^{-3}$. Hence, the inequality is false.

3. Final Answer

B