Question 1064114
Please help me in this equation:
ln(x+1)/1+lnx >0
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It's not an equation, it's an inequality.
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Is it {{{ln(x+1)/1+ln(x) >0 }}} ?
Or {{{ln(x+1)/(1+ln(x)) >0 }}} ?
Parentheses are free, use some.
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And if possible this one too:
(4-x2).ln(2x^2+x)=0
If you mean {{{(4-x^2)*ln(2x^2+x)=0}}}
Use * (Shift 8) for multiply.
{{{(4-x^2)*ln(2x^2+x)=0}}}
4 - x^2 = 0
x = -2, +2
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{{{ln(2x^2+x)=0}}}
{{{ln(2x^2+x) = ln(1)}}}
2x^2 + x = 1
Solve the quadratic for x
Check that the solutions are valid.