How do i solve these problems? ln(x) = 5.6 + ln(7.5) and log(x) = 5.6 - log(7.5)

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

second option

step-by-step explanation:

the measures of a triangle equal 180. use that to find x.

2x+3x+4x=180

9x=180

9x/9=180/9

x=20

now you know x, you know angle a is 40°, angle b is 60°, angle c is 80°.

because angle a is 40°, the first option doesn't work.

because angle b is 60°, the second option does work.

when you add angle a and b, you get 100°. complementary means it equals 90°, so the third option doesn't work.

when you add angle a and c, you get 120°, which is not 100.