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]
1. 12/5
2. 24/10
3. 48/20
4. 96/40
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