SOLUTION: Please help! Solve lnx^2=ln(72-x)
equivalent
e^(lnx^2)=e^[ln(72-x)],because e^x is 1-1 function.
equivalent
x^2=72-x
equivalent
x^2+x-72=0
equivalent
x1=[-1+sqr(1+4*1*72)]
Algebra.Com
Question 386888: Please help! Solve lnx^2=ln(72-x)
equivalent
e^(lnx^2)=e^[ln(72-x)],because e^x is 1-1 function.
equivalent
x^2=72-x
equivalent
x^2+x-72=0
equivalent
x1=[-1+sqr(1+4*1*72)]/2 , x2=[-1-sqr(1+4*1*72)]/2
equivalent
x1=[-1+sqr(1+288)]/2, x2=[-1-sqr(1+288)]/2
equivalent
x1=(-1+17)/2 , x2=(-1-17)/2
equivalent
x1=16/2,x2=-18/2
equivalent
x1=8,x2=-9
.
Found 2 solutions by robertb, dnanos:
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Since ln is a 1-to-1 function, then , or .
(x+9)(x-8) = 0;
x = -9, 8 .
Both x-values satisfy the original equation, and so are the final answers.
Answer by dnanos(83) (Show Source): You can put this solution on YOUR website!
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