Question

1)㏒₂x+6㏒₄x=8 2)(1/3)^x+2 +5 *(1/3)^x+1>16

Answer 1

Смотри......................

Answer 2

To solve the equations, I'll tackle them one by one:

1. ㏒₂x + 6㏒₄x = 8

To simplify this equation, we can use logarithmic rules. Remember that ㏒ₐb = logₐb.

Using the logarithmic rule logₐb + logₐc = logₐ(bc), we can rewrite the equation as:

㏒₂x + ㏒₄x^6 = 8

Now, we can use another logarithmic rule: logₐb^n = n * logₐb. Applying this, the equation becomes:

㏒₂x + 6 * ㏒₂x = 8

Now, we combine the logarithmic terms:

7 * ㏒₂x = 8

Finally, we isolate x:

㏒₂x = 8 / 7

Now, we can rewrite this equation in exponential form:

x = 2^(8 / 7)

Let's calculate the approximate value of x:

x ≈ 2^(8 / 7) ≈ 2.244

So, the solution to the equation is x ≈ 2.244.

2. (1/3)^x+2 + 5 * (1/3)^x+1 > 16

To solve this inequality, we'll first deal with the exponents.

Let's consider (1/3)^x+2:

(1/3)^x+2 = (1/3)^2 * (1/3)^x = (1/9) * (1/3)^x

Now, we'll deal with (1/3)^x+1:

(1/3)^x+1 = (1/3)^1 * (1/3)^x = (1/3) * (1/3)^x = (1/3)^x / 3

Now, we can rewrite the inequality as:

(1/9) * (1/3)^x + 5 * ((1/3)^x / 3) > 16

To make it easier to work with, let's use a substitution:

Let y = (1/3)^x

The inequality becomes:

(1/9) * y + 5 * (y / 3) > 16

Now, we'll simplify:

(1/9) * y + (5/3) * y > 16

To combine the terms, find a common denominator, which is 9:

(1/9) * y + (5/3) * y = (1/9) * y + (15/9) * y = (1/9 + 15/9) * y = (16/9) * y

The inequality now becomes:

(16/9) * y > 16

Now, isolate y:

y > 16 / (16/9)

y > 9

Now, remember that y = (1/3)^x. So, we have:

(1/3)^x > 9

To get rid of the exponent, take the logarithm (base 1/3) of both sides:

㏒_(1/3) ((1/3)^x) > ㏒_(1/3)(9)

Simplify:

x > ㏒_(1/3)(9)

Now, calculate the approximate value:

x > ㏒_(1/3)(9) ≈ 2.727

So, the solution to the inequality is x > 2.727.