Find x given that:
i) e(3x+1) = 5
ii) log3x+ log9x= 6
Maths logarithm problem?
2010-01-22 10:51 am
回答 (4)
2010-01-22 1:57 pm
✔ 最佳答案
i)e(3x + 1) = 5
3x + 1 = 5/e
3x = 5/e - 1
x = (5/e - e/e)/3
x = [(5 - e)/e]/3
x = [(5 - e)/e][1/3]
x = (5 - e)/(3e)
ii)
log_3(x) + log_9(x) = 6
log_3(x) + log_(3^2)(x) = 6
log_3(x) + [log_3(x)]/2 = 6
2log_3(x) + log_3(x) = 12
log_3(x^2) + log_3(x) = 12
log_3(x^3) = 12
x^3 = 3^12
x = 3^(12/3)
x = 3^4
x = 81
2010-01-22 7:00 pm
i) da answer is ln( cube root of 5/e) simply take natural log ln on both sides of equation , put 1 = ln(e) nd get da answer
ii) 1000 divided by 3 root 3 log ab = log a + log b
ii) 1000 divided by 3 root 3 log ab = log a + log b
2010-01-22 6:59 pm
i) (3x+1)=5/e , 3x=5/e-1 , x=(5/e -1)/3=-0.152
ii) log 27x^2=log1000000, x^2=1000000/27, x=1000/27^1/2
God bless you.
ii) log 27x^2=log1000000, x^2=1000000/27, x=1000/27^1/2
God bless you.
2010-01-22 6:55 pm
e^(3x+1) = 5
Take the natural logs of both sides:
ln(e^(3x+1)) = ln5
3x+1 = ln5
3x = ln5 - 1
x = (ln5 - 1) / 3 (0.203146 approx)
ii) log3x + log9x = 6
Then log27x^2 = 6
27x^2 = e^6
x^2 = (e^6)/27
x = sqrt(e^6/27) [take +ve square root as x > 0]
x = (e^3)/(3sqrt(3)) , or approx 3.86546338314
Take the natural logs of both sides:
ln(e^(3x+1)) = ln5
3x+1 = ln5
3x = ln5 - 1
x = (ln5 - 1) / 3 (0.203146 approx)
ii) log3x + log9x = 6
Then log27x^2 = 6
27x^2 = e^6
x^2 = (e^6)/27
x = sqrt(e^6/27) [take +ve square root as x > 0]
x = (e^3)/(3sqrt(3)) , or approx 3.86546338314
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