Question

3*4^x + 2*9^x = 5*6^x

Answer 1

$3 \times ( {2}^{x} ) ^{2} + 2 \times ({3}^{x}) ^{2}- \\ - 5 \times {3}^{x} \times {2}^{x} = 0$
разделим обе части на
${3}^{x} ≠0$

$3 ((\frac{2}{3}) ^{x} )^{2} + 2 - 5( \frac{2}{3} )^{x} = 0$

сделаем замену
$y = ( \frac{2}{3} )^{x} > 0$
3у²-5у+2=0

D=(-5)²-4•3•2=1>0

у¹'²=(5±1)/6
у¹=1
у²=⅔

сделаем обратную замену

$( \frac{2}{3} )^{x} = 1 = > x_1 = 0$
$( \frac{2}{3} )^{x} = \frac{2}{3} = >x _2= 1$


Ответ
$x_1=0; \\ x_2=1$

Answer 2

To solve the equation 34^x + 29^x = 5*6^x, we can try to simplify it using some algebraic manipulation.

First, notice that we can write 6 as 2*3 and 9 as 3^2. So we can rewrite the equation as:

34^x + 2(3^2)^x = 5*(2*3)^x

Simplifying further, we get:

34^x + 23^(2x) = 52^x3^x

Now we can try to isolate the variable x on one side of the equation. Let's start by dividing both sides by 3^x:

3^(x)34^x/3^x + 23^(2x)/3^x = 52^x*3^x/3^x

Simplifying, we get:

34^x + 23^x = 5*2^x

Next, we can divide both sides by 2^x:

3*(4/2)^x + 2*(3/2)^x = 5

Simplifying, we get:

32^x + 2(3/2)^x = 5

Let y = (3/2)^x. Then we can rewrite the equation as:

32^x + 2y = 5

Now we have a linear equation in two variables, x and y. Solving for y, we get:

y = (5 - 3*2^x)/2

Substituting this back into the equation y = (3/2)^x, we get:

(3/2)^x = (5 - 3*2^x)/2

Multiplying both sides by 2, we get:

3^x/2^(x+1) = 5/2 - 3/2*2^x

Multiplying both sides by 2^(x+1), we get:

3^x = 52^(x+1) - 32^x

Simplifying, we get:

3^x = 22^(x+1) + 32^x

3^x = 2^(x+2) + 3*2^x

Now we can try to solve for x. Let's rewrite the equation as:

3^x - 3*2^x = 2^(x+2)

We can see that the left-hand side is a difference of two exponential terms with different bases. To solve for x, we can use a numerical method such as graphing or Newton's method. However, there is no exact algebraic solution for x in this equation.