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


Ответы на вопрос

1)log2(x+1)=2
2^2=x+1
x=4+1
x=5
Всё что смогла(



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


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