Use the rules of logarithms and the rules of exponents.
... ln(ab) = ln(a) + ln(b)
... e^ln(a) = a
... (a^b)·(a^c) = a^(b+c)
_____
1) Use the second rule and take the antilog.
... e^ln(x) = x = e^(5.6 + ln(7.5))
... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents
... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms
... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)
2) Similar to the previous problem, except base-10 logs are involved.
... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.
... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5
... x ≈ 53,080.96
Answer:
6.5 = <em>h</em> + 2
<em>h</em> = 4.5
Step-by-step explanation:
We are given <em>h</em> and since she did 2 additional hours of work, we add the 2 hours. and since we are trying to find <em>h</em>, we set the equation equal to 6.5
Answer:
11 dimes, 17 nickels
Step-by-step explanation:
d + n = 28, put one variable by itself on a side, d = 28 - n
.10d + .05n = 1.95
Sub the first equation into the second
.10(28 - n) + .05n = 1.95
2.8 - .10n + .05n = 1.95
2.8 - .05n = 1.95
-.05n = -0.85
n = 17 nickels
d + 17 = 28
d = 11 11 dimes
Answer:
intervals (-3,-1) and (0,+infinity)
Step-by-step explanation:
if f'(x)>0 then f is increasing
