Question

18 points please help pic inserted

Answer 1

Answer:

1). $\text{log}_4(5x^2+2)=\text{log}_4(x + 8)$

2). log(x - 1) + log5x = 2

3). ln(x + 5) = ln(x - 1) + ln(x + 1)

4). $e^{x^2}=e^{4x+5}$

Step-by-step explanation:

1). ln(x + 5) = ln(x - 1) + ln(x + 1)

   ln(x + 5) = ln(x - 1)(x + 1)  [Since ln(a×b) = ln a + lnb]

   (x + 5) = (x- 1)(x + 1)

   x + 5 = x² - 1

   x² - x - 6 = 0

   x² - 3x + 2x - 6 = 0

   x(x - 3) + 2(x - 3) = 0

   (x + 2)(x - 3 ) = 0

   x = -2, 3

   But x = -2 is an extraneous solution.

  Therefore, x = 3 is the only solution.

2). $e^{x^2}=e^{4x+5}$

  x² = 4x +5

  x² - 4x - 5 = 0

  x² - 5x + x - 5 = 0

  x(x - 5) + 1(x - 5) = 0

  (x + 1)(x - 5) = 0

  x = -1, 5

  Therefore, solution set is (-1, 5)

3). $\text{log}_4(5x^2+2)=\text{log}_4(x + 8)$

   5x² + 2 = (x + 8)  

   5x² - x - 6 = 0

   5x² - 6x + 5x - 6 = 0

   x(5x - 6) + 1(5x - 6) = 0

   (x + 1)(5x - 6) = 0

   x = -1, $\frac{6}{5}$

4). log(x - 1) + log5x = 2

    log(x - 1)(5x) = 2

    5x(x - 1) = 10² [if loga = b, $a=10^{b}$]

    5x² - 5x - 100 = 0

    x² - x - 20 = 0

    x² - 5x + 4x - 20 = 0

    x(x - 5) + 4(x - 5) =0

    (x - 5)(x + 4) = 0

    x = -4, 5

But x = -4 is an extraneous solution.

Therefore, x = 5 is the only solution.