Answer:
1). [tex]\text{log}_4(5x^2+2)=\text{log}_4(x + 8)[/tex]
2). log(x - 1) + log5x = 2
3). ln(x + 5) = ln(x - 1) + ln(x + 1)
4). [tex]e^{x^2}=e^{4x+5}[/tex]
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). [tex]e^{x^2}=e^{4x+5}[/tex]
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). [tex]\text{log}_4(5x^2+2)=\text{log}_4(x + 8)[/tex]
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, [tex]\frac{6}{5}[/tex]
4). log(x - 1) + log5x = 2
log(x - 1)(5x) = 2
5x(x - 1) = 10² [if loga = b, [tex]a=10^{b}[/tex]]
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.
