Question

Log5(7+5x)=log5(7-x)+2

Answer 1

log5(7+5x)=log5(7-x)+2

log5(7+5x) - log5(7-x) = 2

log5( (7+5x)/(7-x) ) = 2

5² =  (7+5x)/(7-x)

25(7-x) = (7+5x)

175 - 25х = 7 + 5х

30х = 168

х = 168/30 = 5,6

Answer 2

To solve the equation log5(7+5x) = log5(7-x) + 2, we need to use logarithm properties to simplify the equation. Here are the steps to solve it:

Step 1: Use the property log_b(a) = c if and only if b^c = a. Step 2: Equate the expressions inside the logarithms.

Starting with the given equation:

log5(7+5x) = log5(7-x) + 2

Step 1: Apply the property log_b(a) = c if and only if b^c = a:

5^(log5(7+5x)) = 5^(log5(7-x) + 2)

Step 2: Apply the power rule of logarithms, log_b(a^c) = c*log_b(a):

7 + 5x = 5^(log5(7-x)) * 5^2

Step 3: Simplify the right side of the equation:

Remember that 5^2 = 25:

7 + 5x = 25 * 5^(log5(7-x))

Step 4: Since we have the same base on both sides of the equation (5), we can equate the exponents:

log5(7-x) = 2

Step 5: Apply the inverse of the logarithm function to both sides to solve for (7-x):

7 - x = 5^2

7 - x = 25

Step 6: Solve for x:

x = 7 - 25

x = -18

So, the solution to the equation is x = -18.