Question

Log4(4+7x)=log4(1+5x)+1. x -???

Answer 1

Ограничения:
4+7x >0
1+5x>0

Log4(4+7x)=log4(1+5x)+1
Log4(4+7x)=log4(1+5x)+log4(4)
Log4(4+7x)=log4((1+5x)*4)
4+7x = (1+5x)*4
 4+7x = 4+20x
7x = 20х
х = 0 (если подставить в ограничения, то х= 0 подходит)

Ответ: 0

Answer 2

To solve the equation log4(4+7x) = log4(1+5x) + 1 for x, we'll use logarithm properties to simplify the equation and find the value of x.

Step 1: Apply the logarithm property log_b(a) - log_b(c) = log_b(a/c). log4(4+7x) = log4(1+5x) + 1 log4(4+7x) - log4(1+5x) = 1

Step 2: Apply the logarithm property log_b(a) = log_b(c) if and only if a = c. Since the bases are the same (both are 4), we can equate the arguments inside the logarithms: 4 + 7x = 1 + 5x

Step 3: Solve for x: Subtract 5x from both sides: 7x - 5x = 1 - 4 2x = -3

Divide both sides by 2 to isolate x: x = -3/2

So, the solution for x is x = -3/2.