Question

Log7(6x-18)>log7(2x-2)

Answer 1

$\left \{ {{6x-18\ \textgreater \ 0} \atop {2x-2\ \textgreater \ 0}} \right. \\ \left \{ {{6x\ \textgreater \ 18} \atop {2x\ \textgreater \ 2}} \right. \\ \left \{ {{x\ \textgreater \ 3} \atop {x\ \textgreater \ 1}} \right. \\ x\ \textgreater \ 3\\ log_7(6x-18)\ \textgreater \ log_7(2x-2)\\ 6x-18\ \textgreater \ 2x-2\\ 4x\ \textgreater \ 16\\ x\ \textgreater \ 4$

Ответ x>4

Answer 2

I can help you with logarithmic inequalities. To solve this problem, you need to apply the following steps:

1. Use the property of logarithms that says $\log_a b > \log_a c$ if and only if $b > c$ for any positive base $a$ and any positive numbers $b$ and $c$. 2. Rewrite the inequality as $6x - 18 > 2x - 2$ by removing the logarithms with base 7. 3. Simplify the inequality by subtracting 2x from both sides and adding 18 to both sides. You get $4x > 16$. 4. Divide both sides by 4 to get the final answer: $x > 4$.

You can check your answer by plugging in some values of $x$ that are greater than 4 and some that are less than 4 into the original inequality. You will see that the inequality is true only when $x > 4$.

If you need more help with logarithms, you can use this online calculator or read this tutorial. I hope this was helpful. Have a nice day!