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]
53+n=(58+10)
n=15
emma got 15 marbles from fred
Answer:
8
Step-by-step explanation:
approx. 8
As 510 lies closer to the cube of 8 which is 512
It means that the number of your answer is too big for the screen. You may want to see if you can work the problem out on paper if you need the numbers it isnt showing you. Just make sure your answer matches the answer you got on the calculator. Minus, of course, the numbers it does not show.
Lets find that out writing equations and solving them, a multiple of 1/6 is something like this (1/6)x, so we have
(1/6)x > 3/6
and
<span>(1/6)y < 4/6
</span>lets solve both equations:
<span>(1/6)x > 3/6
</span>x > 6(3/6)
x > 3
<span>(1/6)y < 4/6
</span>y < 6(4/6)
y < 4
So the number must be between 3 and 4, which is obvious, lets try with 3.5 then, that is 3 5/10 or 35/10 = 7/2 in fractional form, and lets try it out:
(1/6)(7/2) = 7/12
finally we compare with the original fractions:
1/6 < 7/12
2/12 < 7/12
So, it comply with being greater than 1/6, now lets compare with 4/6
7/12 < 4/6
<span>7/12 < 8/12
</span>therefore is also smaller than 4/6 and hence 7/12 is a multiple of 1/6 between 3/6 and 4/6
