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Logarithm of equation e^(2x-1) = 2e^x + 3e x=? Exact form (non decimal)
The original formula is equivalent, lne ^ (2x-1) = ln2e ^ x + ln3e
It is reduced to 2x-1 = x + LN2 + Ln3 + 1
The solution is x = LN2 + Ln3 + 2
x=ln6 +2
Logarithmic equation
1.2lgx+lg7=lg14
2.lgx+lg(x-3)=1
3.lg(x^2+11x+8)-lg(x+1)=1
4.lg(x+6)-1/2lg(2x-3)=2-lg25
1.2lgx+lg7=lgx^2+lg7=lg7*x^2∵2lgx+lg7=lg14∴lg7*x^2=lg14∴7*x^2=14 ∴x=√22.lgx+lg(x-3)=lgx（x-3) ∵1=lg10lgx+lg(x-3)=1∴x（x-3）=10∴x=53.lg(x^2+11x+8)-lg(x+1)=lg(x^2+11x+8)/(x+1)∵1=lg10∴(x^2+11x+...
lg(x^2-8x+9)=lg(3-x)
lg(x^2-8x+9)=lg(3-x)
x^2-8x+9=3-x
x^2-7x+6=0
(x-6)(x-1)=0
X = 6 or x = 1, test X = 6,3-6 = - 3, do not conform, give up
X = 1,3-x = 2
x=1
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