Question

1. (2/7)^х=7/2 2. 2^2х - 3*2^х-4=0

Answer 1

Вот, держи. Вроде, все просто

Answer 2

1. (2/7)^x=7/2. (2/7)^x=(2/7)^(-1). x=-1.
2. 2^(2x)-3•2^x-4=0.
2^x=t. 2^(2x)=t².
t²-3t-4=0.
t²+t-4t-4=0.
t(t+1)-4(t+1)=0.
(t+1)(t-4)=0.
2^x=-1. ☞ ∅
2^x=2². x=2.

Answer 3

To solve these equations, we'll work on them one by one:

1. (2/7)^x = 7/2

To solve for x, we'll first take the logarithm of both sides with base 2/7 (you can use any logarithm base, but this is common):

log((2/7)^x) = log(7/2)

Now, we can use the logarithm properties to simplify the left side:

x * log(2/7) = log(7/2)

Next, isolate x by dividing both sides by log(2/7):

x = log(7/2) / log(2/7)

Use a calculator to compute the value:

x ≈ 3.086

2. 2^(2x) - 3 * 2^x - 4 = 0

To solve this equation, we can use a substitution to simplify the expression. Let y = 2^x. Now the equation becomes:

y^2 - 3y - 4 = 0

Now we can factor the quadratic equation:

(y - 4)(y + 1) = 0

Now set each factor to zero and solve for y:

y - 4 = 0 => y = 4 y + 1 = 0 => y = -1

Now, remember that y = 2^x, so we have:

1. 2^x = 4
2. 2^x = -1

The second equation has no real solutions since 2^x is always positive for any real value of x. However, the first equation can be solved:

2^x = 4

Now, rewrite 4 as 2^2:

2^x = 2^2

Since the bases are the same, we can equate the exponents:

x = 2

So the solutions to the second equation are:

x = 2

In summary, the solutions to the given equations are approximately:

1. x ≈ 3.086
2. x = 2