Solve for y in terms of x: 2(log y - 3log x) = 3?
回答 (6)
2009-04-24 7:56 am
✔ 最佳答案
=> 2(log y - 3log x) = 3log y - 3log x = 3/2
log y = 3/2 + 3 log x
Assuming that log's base is 10:
log y = (3/2) log 10 + 3 log x
log y = log 10^ (3/2) + log x³
log y = log [ 10^ (3/2) x ( x³ ) ]
removing log from both sides;
y = 10 ^ (3/2) x ( x³ )
2009-04-24 11:47 am
2[log(y) - 3log(x)] = 3
log(y) - 3log(x) = 3/2
log(y) = 3/2 + 3log(x)
y = 10^[3/2 + 3log(x)]
y = 10^(3/2) * 10^[3log(x)]
y = 10^(3/2) * 3log(x)
log(y) - 3log(x) = 3/2
log(y) = 3/2 + 3log(x)
y = 10^[3/2 + 3log(x)]
y = 10^(3/2) * 10^[3log(x)]
y = 10^(3/2) * 3log(x)
2009-04-24 7:54 am
log(y/x^3)=3/2=log(31.6227766)
y=31.6227766x^3=(3.16227766x)^3
y=31.6227766x^3=(3.16227766x)^3
2009-04-24 7:51 am
Alog - Anti-log.
2(log y - 3log x) = 3
log y - 3log x = 3/2
log y - log x³ = 3/2
y / x³ = Alog 3/2
y = x³ Alog 3/2
2(log y - 3log x) = 3
log y - 3log x = 3/2
log y - log x³ = 3/2
y / x³ = Alog 3/2
y = x³ Alog 3/2
2009-04-24 7:45 am
log y - 3 log x = 3/2
log y + log x^(-3) = 3/2
log (y / x³) = 3/2
y / x ² = 2^(3/2) ------assuming logs to base 2
y = 2â2 x ²
Note
Any log base may be used.
log y + log x^(-3) = 3/2
log (y / x³) = 3/2
y / x ² = 2^(3/2) ------assuming logs to base 2
y = 2â2 x ²
Note
Any log base may be used.
2009-04-24 7:42 am
logy = 3/2 + 3logx = log(10^(3/2)x^3)--->y = 10^(3/2) x^3
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