Question

625^x-25^(2x-1)>7

Answer 1

625^x-25^(2x-1)>7
625 мы можем приставить как 25^2x, а все остально переписать!
25^2x-25^(2x-1)>7
Мы видим, что основы похожи, поэтому мы их выбрасывавем и пишем так
2x-(2x-1)>7
2x-2x+1>7

2x-2x>7-1
0>6
или

2x-2x+1>7
1>7

 

Answer 2

To solve the inequality $625^x - 25^{2x-1} > 7$, we'll start by rewriting the numbers in terms of a common base. Both 625 and 25 can be expressed as powers of 5:

$625 = 5^4$ and $25 = 5^2$.

So, we have:

$(5^4)^x - (5^2)^{2x-1} > 7$

$5^{4x} - 5^{4x-2} > 7$

Next, let's factor out $5^{4x-2}$:

$5^{4x-2} \cdot (5^2 - 1) > 7$

$5^{4x-2} \cdot 24 > 7$

$5^{4x-2} > \frac{7}{24}$

Now, take the logarithm of both sides to solve for x:

$4x - 2 > \log_5\left(\frac{7}{24}\right)$

$4x > 2 + \log_5\left(\frac{7}{24}\right)$

$x > \frac{2 + \log_5\left(\frac{7}{24}\right)}{4}$

This is the solution to the inequality $625^x - 25^{2x-1} > 7$. Keep in mind that $x$ must also satisfy any domain restrictions that might apply.