Question

Log2(x+1)=2   log17(5x+7)=log17^22   log3(7x+1)=3log9^4   log3(4+x) =log3^4

Answer 1

1)log2(x+1)=2

2^2=x+1

x=4+1

x=5

 

Всё что смогла(

 

 

 

 

Answer 2

To solve the given equations, we'll use logarithmic properties and algebraic manipulations. Let's solve them one by one:

1. log2(x+1) = 2

Using the property of logarithms, we can rewrite the equation as an exponential equation:

2^(log2(x+1)) = 2^2

(x+1) = 4

x = 4 - 1

x = 3

So the solution to the first equation is x = 3.

2. log17(5x+7) = log17^22

Using the property of logarithms, we can equate the bases:

5x + 7 = 22

5x = 22 - 7

5x = 15

x = 15/5

x = 3

So the solution to the second equation is x = 3.

3. log3(7x+1) = 3log9^4

Using the property of logarithms, we can rewrite the equation as:

log3(7x+1) = log3((9^4)^3)

log3(7x+1) = log3(9^12)

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

7x + 1 = 9^12

7x = 9^12 - 1

x = (9^12 - 1)/7

This is the exact solution for the equation.

4. log3(4+x) = log3^4

Using the property of logarithms, we can rewrite the equation as:

4 + x = 4

x = 0

So the solution to the fourth equation is x = 0.

Summary of solutions:

• The first equation, log2(x+1) = 2, has a solution of x = 3.
• The second equation, log17(5x+7) = log17^22, has a solution of x = 3.
• The third equation, log3(7x+1) = 3log9^4, has an exact solution of x = (9^12 - 1)/7.
• The fourth equation, log3(4+x) = log3^4, has a solution of x = 0.