Question

5*4^x -7*10^x +2*25^x=0

Answer 1

$5*4^x-7*10^x+2*25^x=0\\5*2^{2x}-7*2^x*5^x+2*5^{2x}=0|:2^{2x}\\5-7*(5/2)^x+2*(5/2)^{2x}=0\\t=(5/2)^x\\5-7t+2t^2=0\\D=(-7)^2-4*2*5=49-40=9=3^2\\t_1=(7+3)/4=10/4=5/2\\t_2=(7-3)/4=4/4=1\\\\(5/2)^x=5/2\\(5/2)^x=(5/2)^1\\x=1\\\\(5/2)^x=1\\(5/2)^x=(5/2)^0\\x=0$

Ответ: 0; 1

Answer 2

To solve this equation, we can notice that each term contains a power of a different base (4, 10, and 25), which makes it difficult to solve directly. However, we can use a change of variable to simplify the equation.

Let's substitute y = 2^x. Then, we have:

5y^2 - 710^(log10(y)/log10(2)) + 2*25^(log25(y)/log5(2)) = 0

Notice that we can use logarithmic identities to simplify the exponents in the second and third terms:

10^(log10(y)/log10(2)) = 2^(log2(10)log10(y)/log2(10)) = y^(log2(10)) 25^(log25(y)/log5(2)) = (5^2)^(log5(y)/log5(2)) = y^(2log5(5))

Substituting these into the equation, we get:

5y^2 - 7y^(log2(10)) + 2y^(2log5(5)) = 0

Now, let's substitute z = y^(log2(10)). Then, we have:

y^2 = z^(log10(2)) y^(2log5(5)) = z^(2log10(5))

Substituting these into the equation, we get:

5z^(2log10(2)) - 7z + 2z^(2*log10(5)) = 0

This is a quadratic equation in z, which we can solve using the quadratic formula:

z = [7 ± sqrt(7^2 - 452^(2log10(5)))] / (25)

z = [7 ± sqrt(49 - 40*log10(5))] / 10

z ≈ 0.704 or z ≈ 1.448

Now, we can substitute back for y and solve for x:

y = z^(1/log2(10))

x = log2(y) = log2(z^(1/log2(10))) = (1/log2(10))*log2(z)

x ≈ -0.398 or x ≈ 0.654

Therefore, the solutions to the equation 54^x -710^x +2*25^x=0 are approximately x = -0.398 and x = 0.654.