Question

1)log1/4(2x–7)=–1 2)6^х+6x+1=42 3)4x=64 4)4x=5 5)log2(x2–x–12)=3 6)lgx+lg x–6=0

Answer 1

Log1/4(2x-7)=-1
log1/4(2x-7)=-1, x>7/2
2x-7=(1/4)^-1
2x-7=0,25^-1
2x-7=(1/4)^1
2x-7=4
2x=4+7
2x=11
х=11/2 ; х >7/2
3)4х=64
х=64÷4
х=16

Answer 2

Let's solve each of these equations step by step:

1. log₁/₄(2x - 7) = -1

To solve this equation, we can use the definition of logarithms. Recall that logₐ(b) = c is equivalent to aᶜ = b.

So, in this case, we have:

1/₄⁻¹ = 2x - 7

Now, solve for 2x - 7:

2x - 7 = 4

Add 7 to both sides:

2x = 4 + 7 2x = 11

Divide by 2:

x = 11/2

2. 6^x + 6x + 1 = 42

Subtract 1 from both sides:

6^x + 6x = 41

This equation doesn't have a simple algebraic solution, so you might need to use numerical methods or a calculator to approximate the value of x.

3. 4x = 64

Divide both sides by 4:

x = 64 / 4 x = 16

4. 4x = 5

Divide both sides by 4:

x = 5 / 4

5. log₂(x² - x - 12) = 3

To solve for x, you can use the definition of logarithms:

2³ = x² - x - 12

8 = x² - x - 12

Add 12 to both sides:

x² - x = 20

Now, rearrange the equation and set it equal to zero:

x² - x - 20 = 0

Now, factor the quadratic:

(x - 5)(x + 4) = 0

Set each factor equal to zero and solve for x:

x - 5 = 0 x = 5

x + 4 = 0 x = -4

So, there are two possible solutions: x = 5 and x = -4.

6. lg(x) + lg(x - 6) = 0

It appears you are using "lg" to represent the base-10 logarithm, commonly denoted as "log₁₀."

Combine the logarithms:

log₁₀(x) + log₁₀(x - 6) = 0

Now, you can use the properties of logarithms to combine these into a single logarithm:

log₁₀(x * (x - 6)) = 0

Now, use the fact that log₁₀(1) = 0:

x * (x - 6) = 1

x² - 6x - 1 = 0

This is a quadratic equation. You can solve it using the quadratic formula:

x = [6 ± √(6² - 4 * 1 * (-1))] / (2 * 1)

x = [6 ± √(36 + 4)] / 2

x = [6 ± √40] / 2

x = [6 ± 2√10] / 2

Simplify:

x = 3 ± √10

So, there are two possible solutions: x = 3 + √10 and x = 3 - √10.