Logarithm?

2016-11-09 1:24 pm
1.Solve (5^2x+2) +7(5^x)-8=?
2.Solve (e^2x+2)+27(e^x)-7=?

3.solve (2^2x) (3^3y)=41.4 and (5^5x) (7^7y)=43.8 x=? y=?
solve 5(2^x)=3^y and x^2+y=28.4
x1=?
x2=?
y1=?
y2=?
更新1:

1.Solve (5^2x+2) +7(5^x)-8=0 2.Solve (e^2x+2)+27(e^x)-7=0

回答 (1)

2016-11-09 2:36 pm
✔ 最佳答案
1.
(5^2x+2) +7(5^x)-8=0
25(5^2x)+7(5^x)-8=0
5^x=0.442752091 or 5^x=0.722752091(rej.)
x=-0.506229704

2.
(e^2x+2)+27(e^x)-7=0
e^x=0.243087723 or e^x=-3.897140371(rej.)
x=-1.14332897

3
(2^2x) (3^3y)=41.4
(x log 4)+(y log 27)=log 41.4 i
(5^5x) (7^7y)=43.8
(x log 3125)+(y log 823543)=log 43.8 ii
x=-5.008399418
y=3.236324244

4
5(2^x)=3^y
(log 5)+(x log 2)=y log 3
(x log 2)-(y log 3)=-log 5 i
x^2+y=28.4 ii
x1=4.884010536
x2=-5.51494029
y1=4.546441085
y2=-2.014566397


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