Question

1)Log3(3x+1)>2 2)Log1/7(4x+1)<-2​

Answer 1

Ответ:

1) x>2 2/3

2) x > 12

Объяснение:

1)

${3}^{log3(3x + 1)} > {3}^{2} \\ 3x + 1 > 9 \\ x > 2\frac{2}{3}$

$2) { \frac{1}{7} }^{log \frac{1}{7} (4x + 1)} > { \frac{1}{7} }^{ - 2} \\ 4x + 1 > 49 \\ x > 12$ неравенство меняется на обратное, т. к. основание 1/7 < 1

Answer 2

To solve the given inequalities involving logarithms, you'll need to isolate the variable $x$. Here's how you can do it for each inequality:

1. $\log_3(3x+1) > 2$

First, convert the logarithmic equation to its exponential form:

$3^2 < 3x+1$

Simplify:

$9 < 3x+1$

Now, subtract 1 from both sides:

$8 < 3x$

Finally, divide both sides by 3:

$\frac{8}{3} < x$

So, the solution to the first inequality is $x > \frac{8}{3}$.

2. $\log_{\frac{1}{7}}(4x+1) < -2$

Again, convert the logarithmic equation to its exponential form:

$\left(\frac{1}{7}\right)^{-2} > 4x+1$

Simplify the left side:

$49 > 4x+1$

Now, subtract 1 from both sides:

$48 > 4x$

Finally, divide both sides by 4:

$12 > x$

So, the solution to the second inequality is $x < 12$.

Therefore, the solutions to the given inequalities are:

1. $x > \frac{8}{3}$
2. $x < 12$