Question

^ - Степень 1) 1,3^3x-2=3 2) 25^x+2·5^x-15=0 3)16^x-4^x+1-14=0

Answer 1

1) (13/10)2*x-2=3, 169/100x-2=3, 100*169/100x-100*2=100*3, 169x-200=300,  169x=300+200,  169x=500,   x=500/169.

2)625x+2*25-15=0,  625x+50x-15=0,  675x-15=0675x=15,  x=15/675,  x=1/45

3)265x-16+1-14=0,  240X=13,  X=13/240

Answer 2

To solve these exponential equations, we'll use some properties of exponents. Let's go through each equation step by step:

1. $1.3^{3x-2} = 3$

Step 1: Take the logarithm (log) of both sides. You can use any base for the logarithm, but let's use the natural logarithm (ln) here:

$ln(1.3^{3x-2}) = ln(3)$

Step 2: Apply the property of logarithms $\log_a(b^c) = c \cdot \log_a(b)$:

$(3x-2) \cdot ln(1.3) = ln(3)$

Step 3: Solve for $x$:

$3x - 2 = \frac{ln(3)}{ln(1.3)}$

$x = \frac{ln(3)}{3 \cdot ln(1.3)}$

Use a calculator to find the approximate value of $x$.

2. $25^{x+2} \cdot 5^{x-15} = 0$

We have a product equal to zero. To get zero, at least one of the factors must be zero. However, $25^{x+2}$ and $5^{x-15}$ are always positive (since any positive number raised to any power is positive), so their product can never be zero. Therefore, there is no real solution for this equation.

3. $16^x - 4^x + 1 - 14 = 0$

Step 1: Combine like terms:

$16^x - 4^x - 13 = 0$

Step 2: Notice that both $16$ and $4$ can be written as powers of $2$:

$(2^4)^x - (2^2)^x - 13 = 0$

Step 3: Apply the properties of exponents $(a^m)^n = a^{m \cdot n}$:

$2^{4x} - 2^{2x} - 13 = 0$

Step 4: Let $u = 2^x$, then the equation becomes:

$u^4 - u^2 - 13 = 0$

This is now a quadratic equation in $u^2$.

Step 5: Solve for $u^2$:

$u^2 = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-13)}}{2 \cdot 1}$

$u^2 = \frac{1 \pm \sqrt{1 + 52}}{2}$

$u^2 = \frac{1 \pm \sqrt{53}}{2}$

Step 6: Find the two possible values of $u$ (positive and negative square roots):

$u_1^2 = \frac{1 + \sqrt{53}}{2}$

$u_2^2 = \frac{1 - \sqrt{53}}{2}$

Step 7: Solve for $x$ using the values of $u$:

For $u_1^2 = \frac{1 + \sqrt{53}}{2}$:

$2^x = \frac{1 + \sqrt{53}}{2}$