Respuesta :
Answer:
(a) ln(x) = 0
Then 0 < x < 1
(b) e^x > 2
Then ln2 < x < ∞
(a) ln(3x - 17) = 5
x = 55.1377197
ln(a + b) + ln(a - b) - 5ln(c)
= ln[(a² - b²)/c^5]
Step-by-step explanation:
First Part.
(a) ln(x) < 0
=> x < e^(0)
x < 1 ....................................(1)
But the logarithm of 0 is 1, and the logarithm of negative numbers are undefined, we can exclude the values of x ≤ 0.
In fact the values of x that satisfy this inequalities are between 0 and 1.
Therefore, we write:
0 < x < 1
(b) e^x > 2
This means x > ln2
and must be finite.
We write as:
ln2 < x < ∞
Second Part.
(a) ln(3x - 17) = 5
3x - 17 = e^5
3x = 17 + e^5
x = (1/3)(17 + e^5)
= 55.1377197
Third Part.
We need to write
ln(a + b) + ln(a - b) - 5ln(c)
as a single logarithm.
ln(a + b) + ln(a - b) - 5ln(c)
= ln(a + b) + ln(a - b) - ln(c^5)
= ln[(a + b)(a - b)/(c^5)]
= ln[(a² - b²)/c^5]
Using exponential and logarithmic functions, it is found that:
a) The solution of the inequality [tex]\ln{(x)} < 0[/tex] is [tex](-infty, 1)[/tex].
b) The solution of the inequality [tex]e^x > 2[/tex] is [tex](\ln{2}, \infty)[/tex]
a) The solution to the equation [tex]\ln{(3x - 17)} = 5[/tex] is x = 55.14.
As a single logarithm, the expression is:
[tex]\ln{\left(\frac{a^2 - b^2}{c^5}\right)}[/tex]
Inequality a:
[tex]\ln{(x)} < 0[/tex]
Applying the exponential to both sides:
[tex]e^{\ln{(x)}} < e^{0}[/tex]
[tex]x < 1[/tex]
Hence, in interval notation:
[tex](-infty, 1)[/tex]
Inequality b:
[tex]e^x > 2[/tex]
Applying ln to both sides:
[tex]\ln{e^x} > \ln{2}[/tex]
[tex]x > \ln{2}[/tex]
Hence, in interval notation, the solution is:
[tex](\ln{2}, \infty)[/tex]
Equation a:
[tex]\ln{(3x - 17)} = 5[/tex]
[tex]e^{\ln{(3x - 17)}} = e^5[/tex]
[tex]3x - 17 = e^5[/tex]
[tex]3x = 17 + e^5[/tex]
[tex]x = \frac{17 + e^5}{3}[/tex]
[tex]x = 55.14[/tex]
The quantity given is:
[tex]\ln{(a + b)} + \ln{(a - b)} - 5\ln{c}[/tex]
To express as a single logarithm, these following properties are applied:
[tex]\ln{a} + \ln{b} = \ln{ab}[/tex]
[tex]\ln{a} - \ln{b} = \ln{\left(\frac{a}{b}\right)}[/tex]
[tex]a\ln{b} = \ln{b^a}[/tex]
Hence:
[tex]\ln{(a + b)} + \ln{(a - b)} - 5\ln{c} = \ln{(a + b)(a - b)} - \ln{c^5}[/tex]
As a single logarithm, the expression is:
[tex]\ln{\left(\frac{a^2 - b^2}{c^5}\right)}[/tex]
A similar problem is given at https://brainly.com/question/21506771
