im trying tosolve e^x=12 and 11^-6y=5 for the first part i got 2.8 but im not sure how to solve for y, please help!?
回答 (3)
2016-06-15 1:52 pm
e^x = 12
ln(e^x) = ln(12)
x ln(e) = ln(12)
x = ln(12)
====
11^(-6y) = 5
log[11^(-6y)] = log(5)
-6y log(11) = log(5)
y = -log(5)/[6log(11)]
ln(e^x) = ln(12)
x ln(e) = ln(12)
x = ln(12)
====
11^(-6y) = 5
log[11^(-6y)] = log(5)
-6y log(11) = log(5)
y = -log(5)/[6log(11)]
2016-06-15 2:09 pm
e^(x) = 12
Ln[e^(x)] = Ln(12)
x.Ln(e) = Ln(12)
x = Ln(12)
x ≈ 2.485
11^(- 6y) = 5
Ln[11^(- 6y)] = Ln(5)
- 6y.Ln(11) = Ln(5)
- 6y = Ln(5)/Ln(11)
y = - (1/6).Ln(5)/Ln(11)
y ≈ - 0.11186
Ln[e^(x)] = Ln(12)
x.Ln(e) = Ln(12)
x = Ln(12)
x ≈ 2.485
11^(- 6y) = 5
Ln[11^(- 6y)] = Ln(5)
- 6y.Ln(11) = Ln(5)
- 6y = Ln(5)/Ln(11)
y = - (1/6).Ln(5)/Ln(11)
y ≈ - 0.11186
2016-06-15 1:56 pm
I'm assuming you meant 11^(-6y) = 5.
11^(-6y) = 5
log₁₁(11^(-6y)) = log₁₁(5)
-6y = log₁₁(5)
y = -⅙log₁₁5
= log₁₁(1/5^⅙)
≅ -0.1119
11^(-6y) = 5
log₁₁(11^(-6y)) = log₁₁(5)
-6y = log₁₁(5)
y = -⅙log₁₁5
= log₁₁(1/5^⅙)
≅ -0.1119
收錄日期: 2021-04-18 15:06:01
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