Question

Log_12(2x-9)=1log_3(5x+3)= 1+3log_3 2

Answer 1

ответ:

log12(2x-9)=1

log12(2x-9)-log12(12)

2x-9)

log 12( --------)

12

log3(5x+3)=1+3 log3(2)

log3(5x+3)=1+3log3(2), x> - 3/5

log3(5x+3)=log3(3)+log3(2³)

log3(5x+3)=log3(3×2³)

log3(5x+3)=log3(3×8)

log3(5x+3)=log3(24)

5x+3=24

5x=24-3

5x=21

x= 21/5, x>-3/5

x=21/5

Answer 2

To solve this system of logarithmic equations:

1. Start with the first equation:

   bashCopy code
   log₁₂(2x - 9) = 1
   

2. Since the base of the logarithm is 12 and the exponent is 1, you can rewrite this equation in exponential form as:

   Copy code
   2x - 9 = 12¹
   

3. Simplify:

   Copy code
   2x - 9 = 12
   

4. Add 9 to both sides to isolate 2x:

   Copy code
   2x = 12 + 9
   2x = 21
   

5. Divide both sides by 2 to solve for x:

   makefileCopy code
   x = 21 / 2
   x = 10.5
   

Now that you've found the value of x from the first equation, you can substitute it into the second equation:

1. Substitute x = 10.5 into the second equation:

   bashCopy code
   log₃(5x + 3) = 1 + 3 * log₃(2)
   

2. Calculate log₃(2):

   bashCopy code
   log₃(2) ≈ 0.63093
   

3. Multiply 3 by log₃(2):

   bashCopy code
   3 * log₃(2) ≈ 3 * 0.63093 ≈ 1.89279
   

4. Rewrite the equation with the known values:

   bashCopy code
   log₃(5 * 10.5 + 3) = 1 + 1.89279
   

5. Calculate 5 * 10.5 + 3:

   Copy code
   5 * 10.5 + 3 = 52.5 + 3 = 55.5
   

6. Rewrite the equation:

   bashCopy code
   log₃(55.5) = 1 + 1.89279
   

7. Subtract 1.89279 from both sides to isolate the logarithm:

   bashCopy code
   log₃(55.5) - 1.89279 = 1
   

8. Finally, convert this equation back to exponential form:

   Copy code
   3¹ = 55.5
   

Since 3¹ is equal to 3, the equation is true. So, the solution to the system of equations is x = 10.5.