Question

5^(x+1)-3^(x+2)≥43*5^(x-1)-19*3^x​

Answer 1

Ответ:

Объяснение:

5^(x+1)-3^(x+2)≥43*5^(x-1)-19*3^x​

5*5^(x)-9*3^(x)≥43/5*5^(x)-19*3^x​

(5-43/5)*5^(x))≥(9-19)*3^x​

-(18/5)*5^(x)) ≥ - (10)*3^x​

-(18/5)*(5/3)^(x)) ≥ - (10)

(5/3)^(x) $\leq$ 50/18 = 25/9 = (5/3)^(2)

(x) $\leq$ (2) - это ответ

Answer 2

To solve the inequality $5^{x+1} - 3^{x+2} \geq 43 \cdot 5^{x-1} - 19 \cdot 3^x$, we can first simplify it and then isolate (x:

Step 1: Simplify the equation:

$5^{x+1} - 3^{x+2} \geq 43 \cdot 5^{x-1} - 19 \cdot 3^x$

Now, let's rewrite $5^{x+1}$ as $5 \cdot 5^x$ and $3^{x+2}$ as $9 \cdot 3^x$:

$5 \cdot 5^x - 9 \cdot 3^x \geq 43 \cdot 5^{x-1} - 19 \cdot 3^x$

Step 2: Rearrange the terms to isolate $x$ on one side of the inequality:

Move all the terms involving $x$ to the left side and the constants to the right side:

$5 \cdot 5^x - 9 \cdot 3^x + 19 \cdot 3^x \geq 43 \cdot 5^{x-1}$

Combine like terms on the left side:

$(5 - 9 + 19) \cdot 3^x \geq 43 \cdot 5^{x-1}$

$15 \cdot 3^x \geq 43 \cdot 5^{x-1}$

Step 3: Now, we can isolate $x$ by dividing both sides by $15 \cdot 3^x$ and $5^{x-1}$:

$\frac{15 \cdot 3^x}{15 \cdot 3^x} \geq \frac{43 \cdot 5^{x-1}}{15 \cdot 3^x}$

Simplify:

$1 \geq \frac{43}{15} \cdot \frac{5^{x-1}}{3^x}$

Now, let's simplify further:

$1 \geq \frac{43}{15} \cdot \frac{5^{x-1}}{3^x}$

Now, you can take the natural logarithm (ln) of both sides to solve for $x$:

$ln(1) \geq ln\left(\frac{43}{15} \cdot \frac{5^{x-1}}{3^x}\right)$

$0 \geq ln\left(\frac{43}{15} \cdot \frac{5^{x-1}}{3^x}\right)$

Now, you can use the properties of logarithms to simplify further. The inequality $ln(1) = 0$, and the logarithm of a product can be expressed as the sum of logarithms:

$0 \geq ln\left(\frac{43}{15}\right) + ln\left(\frac{5^{x-1}}{3^x}\right)$

Now, isolate the second term on the right side:

$0 \geq ln\left(\frac{43}{15}\right) + ln\left(\frac{5^{x-1}}{3^x}\right)$

Use the properties of logarithms to simplify further:

$0 \geq ln\left(\frac{43}{15} \cdot \frac{5^{x-1}}{3^x}\right)$