Question

9×7^x-49×3^x>0 ????????????

Answer 1

9×7^x-49×3^x>0 
3²7ˣ-7²3ˣ>0
3²7ˣ> 7²3ˣ
(7/3)ˣ>(7/3)²
X>2

Answer 2

9=3^2
49=7^2

3^2*7^х - 7^2*3^х >0

7^х - 7^2*3^(х-2) >0 все рпзделить на 3^2

7^(х-2) - 3^(х-2) >0 все разделить на 7^2

7^(х-2) > 3^(х-2)

ln(7^(х-2)) > ln(3^(х-2)) ln - натуральный логарифм

(х-2)*ln7 > (x-2)*ln3

Отсюда ln7>ln3, если коэффициент положительный

х-2>0
х>2
Ответ: (2;+бесконечность)

Answer 3

To solve the inequality 9×7^x - 49×3^x > 0, we can start by factoring out a common term.

Let's express both terms in powers of 3: 9×(7^x) - 49×(3^x)

Now, we can rewrite 7^x as (3^2)^x, which is equal to 3^(2x):

9×(3^(2x)) - 49×(3^x)

To simplify further, we can let y = 3^x:

9y^2 - 49y > 0

Now we have a quadratic inequality. Let's solve it by factoring:

y(9y - 49) > 0

The expression can be positive in two cases:

1. When both y and (9y - 49) are positive.
2. When both y and (9y - 49) are negative.

Let's solve each case separately:

Case 1: y > 0 and 9y - 49 > 0 y > 0 and y > 49/9

Since y = 3^x, we have: 3^x > 0 and 3^x > 49/9

The first inequality, 3^x > 0, is always true since any positive number raised to any power will be positive.

For the second inequality, 3^x > 49/9, we can take the logarithm (base 3) of both sides to solve for x:

x > log3(49/9)

Case 2: y < 0 and 9y - 49 < 0 y < 0 and y < 49/9

Similarly, since y = 3^x, we have: 3^x < 0 and 3^x < 49/9

The first inequality, 3^x < 0, is never true since any positive number raised to any power will be positive.

For the second inequality, 3^x < 49/9, we can take the logarithm (base 3) of both sides to solve for x:

x < log3(49/9)

To summarize the solutions: x > log3(49/9) for the positive case.

Therefore, the solution to the inequality 9×7^x - 49×3^x > 0 is: x > log3(49/9)