Solve each inequality for x. (Enter your answers using interval notation.)

(a) ln(x) < 0

(b) ex > 2

Solve each equation for x.

(a) ln(3x − 17) = 5

Express the given quantity as a single logarithm.

ln(a + b) + ln(a − b) − 5 ln(c)

(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]

4th option is correct xy=-10

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